积分表及公式推导

（一）含有 ax+b\sqrt{ax+b} 的积分

  1. ∫dxax+b=1aln⁡∣ax+b∣+C\displaystyle\int\dfrac{\mathrm{d}x}{ax+b}=\dfrac{1}{a}\ln|ax+b|+C

∫dxax+b=1a∫d(ax+b)ax+b=1aln⁡∣ax+b∣+C\begin{align} \int\dfrac{\mathrm{d}x}{ax+b} & = \dfrac{1}{a}\int\dfrac{\mathrm{d}(ax+b)}{ax+b}\\ & = \dfrac{1}{a}\ln|ax+b|+C \end{align}

  1. ∫(ax+b)μdx=1a(μ+1)(ax+b)μ+1+C(μ≠1)\displaystyle\int(ax+b)^\mu\mathrm{d}x=\dfrac{1}{a(\mu+1)}(ax+b)^{\mu+1}+C\quad(\mu\neq1)

∫(ax+b)μdx=1a∫(ax+b)μd(ax+b)=1a(μ+1)(ax+b)μ+1+C(μ≠1)\begin{align} \int(ax+b)^\mu\mathrm{d}x & =\dfrac{1}{a}\int(ax+b)^\mu\mathrm{d}(ax+b)\\ & = \dfrac{1}{a(\mu+1)}(ax+b)^{\mu+1}+C (\mu\neq1) \end{align}

  1. ∫xax+bdx=1a2(ax+b−bln⁡∣ax+b∣)+C\displaystyle\int\dfrac{x}{ax+b}\mathrm{d}x=\dfrac{1}{a^2}(ax+b-b\ln|ax+b|)+C

∫xax+bdx=∫(1a−ba⋅1ax+b)dx=1ax−ba⋅1aln⁡∣ax+b∣+C′=1a2(ax−bln⁡∣ax+b∣)+C′=1a2(ax+b−bln⁡∣ax+b∣)+C\begin{align} & \int\dfrac{x}{ax+b}\mathrm{d}x\\ = & \int\left(\frac{1}{a}-\frac{b}{a}\cdot\frac{1}{ax+b}\right)\mathrm{d}x\\ = & \frac{1}{a}x-\frac{b}{a}\cdot\frac{1}{a}\ln|ax+b|+C'\\ = & \frac{1}{a^2}(ax-b\ln|ax+b|)+C'\\ = & \frac{1}{a^2}(ax+b-b\ln|ax+b|)+C \end{align}

  1. ∫x2ax+bdx=1a3[12(ax+b)2−2b(ax+b)+b2ln⁡∣ax+b∣]+C\displaystyle\int\dfrac{x^2}{ax+b}\mathrm{d}x=\dfrac{1}{a^3}\left[\dfrac{1}{2}(ax+b)^2-2b(ax+b)+b^2\ln|ax+b|\right]+C

∫x2ax+bdx=∫1a2(ax+b)2−2ba2(ax+b)+b2a2ax+bdx=1a2∫(ax+b)dx−2ba2∫dx+b2a2∫1ax+bdx=1a2(12ax2+bx)−2ba2x+b2a3ln⁡∣ax+b∣+C=1a3[12(ax+b)2−2b(ax+b)+b2ln⁡∣ax+b∣]+C\begin{align} & \int\frac{x^2}{ax+b}\mathrm{d}x\\ = & \int\frac{\frac{1}{a^2}(ax+b)^2-\frac{2b}{a^2}(ax+b)+\frac{b^2}{a^2}}{ax+b}\mathrm{d}x\\ = & \frac{1}{a^2}\int(ax+b)\mathrm{d}x-\frac{2b}{a^2}\int\mathrm{d}x+\frac{b^2}{a^2}\int\frac{1}{ax+b}\mathrm{d}x\\ = & \frac{1}{a^2}\left(\frac{1}{2}ax^2+bx\right)-\frac{2b}{a^2}x+\frac{b^2}{a^3}\ln|ax+b|+C\\ = & \dfrac{1}{a^3}\left[\dfrac{1}{2}(ax+b)^2-2b(ax+b)+b^2\ln|ax+b|\right]+C \end{align}

  1. ∫dxx(ax+b)=−1bln⁡∣ax+bx∣+C\displaystyle\int\dfrac{\mathrm{d}x}{x(ax+b)}=-\dfrac{1}{b}\ln\left|\dfrac{ax+b}{x}\right|+C

∫dxx(ax+b)=∫(1bx−abax+b)dx=1bln⁡∣x∣−1bln⁡∣ax+b∣+C=−1bln⁡∣ax+bx∣+C\begin{align} & \int\dfrac{\mathrm{d}x}{x(ax+b)}\\ = & \int\left(\frac{\frac{1}{b}}{x}-\frac{\frac{a}{b}}{ax+b}\right)\mathrm{d}x\\ = & \frac{1}{b}\ln|x|-\frac{1}{b}\ln|ax+b|+C\\ = & -\frac{1}{b}\ln\left|\frac{ax+b}{x}\right|+C \end{align}

  1. ∫dxx2(ax+b)=−1bx+ab2ln⁡∣ax+bx∣+C\displaystyle\int\dfrac{\mathrm{d}x}{x^2(ax+b)}=-\dfrac{1}{bx}+\dfrac{a}{b^2}\ln\left|\dfrac{ax+b}{x}\right|+C

∫dxx2(ax+b)=∫(1bx2+−ab2x+a2b2ax+b)dx=1b∫1x2dx−ab2∫1xdx+a2b2∫1ax+bdx=−1bx+ab2ln⁡∣ax+bx∣+C\begin{align} & \int\dfrac{\mathrm{d}x}{x^2(ax+b)}\\ = & \int\left(\frac{\frac{1}{b}}{x^2}+\frac{-\frac{a}{b^2}}{x}+\frac{\frac{a^2}{b^2}}{ax+b}\right)\mathrm{d}x\\ = & \frac{1}{b}\int\frac{1}{x^2}\mathrm{d}x-\frac{a}{b^2}\int\frac{1}{x}\mathrm{d}x+\frac{a^2}{b^2}\int\frac{1}{ax+b}\mathrm{d}x\\ = & -\dfrac{1}{bx}+\dfrac{a}{b^2}\ln\left|\dfrac{ax+b}{x}\right|+C \end{align}

  1. ∫x(ax+b)2dx=1a2[ln⁡∣ax+b∣+bax+b]+C\displaystyle\int\dfrac{x}{(ax+b)^2}\mathrm{d}x=\dfrac{1}{a^2}\left[\ln|ax+b|+\dfrac{b}{ax+b}\right]+C

∫x(ax+b)2dx=∫(1aax+b−ba(ax+b)2)dx=1a∫1ax+bdx−ba∫1(ax+b)2dx=1a2[ln⁡∣ax+b∣+bax+b]+C\begin{align} & \int\dfrac{x}{(ax+b)^2}\mathrm{d}x\\ = & \int\left(\frac{\frac{1}{a}}{ax+b}-\frac{\frac{b}{a}}{(ax+b)^2}\right)\mathrm{d}x\\ = & \frac{1}{a}\int\frac{1}{ax+b}\mathrm{d}x-\frac{b}{a}\int\frac{1}{(ax+b)^2}\mathrm{d}x\\ = & \dfrac{1}{a^2}\left[\ln|ax+b|+\dfrac{b}{ax+b}\right]+C \end{align}

  1. ∫x2(ax+b)2dx=1a3(ax+b−2bln⁡∣ax+b∣−b2ax+b)+C\displaystyle\int\dfrac{x^2}{(ax+b)^2}\mathrm{d}x=\dfrac{1}{a^3}\left(ax+b-2b\ln|ax+b|-\dfrac{b^2}{ax+b}\right)+C

∫x2(ax+b)2dx=∫1a2(ax+b)2−2ba2(ax+b)+b2a2(ax+b)2dx=1a2∫dx−2ba2∫1ax+bdx+b2a2∫1(ax+b)2dx=1a3(ax−2bln⁡∣ax+b∣−b2ax+b)+C′=1a3(ax+b−2bln⁡∣ax+b∣−b2ax+b)+C\begin{align} & \int\dfrac{x^2}{(ax+b)^2}\mathrm{d}x\\ = & \int\frac{\frac{1}{a^2}(ax+b)^2-\frac{2b}{a^2}(ax+b)+\frac{b^2}{a^2}}{(ax+b)^2}\mathrm{d}x\\ = & \frac{1}{a^2}\int\mathrm{d}x-\frac{2b}{a^2}\int\frac{1}{ax+b}\mathrm{d}x+\frac{b^2}{a^2}\int\frac{1}{(ax+b)^2}\mathrm{d}x\\ = & \frac{1}{a^3}\left(ax-2b\ln|ax+b|-\frac{b^2}{ax+b}\right)+C'\\ = & \dfrac{1}{a^3}\left(ax+b-2b\ln|ax+b|-\dfrac{b^2}{ax+b}\right)+C \end{align}

  1. ∫dxx(ax+b)2=1b(ax+b)−1b2ln⁡∣ax+bx∣+C\displaystyle\int\dfrac{\mathrm{d}x}{x(ax+b)^2}=\dfrac{1}{b(ax+b)}-\dfrac{1}{b^2}\ln\left|\dfrac{ax+b}{x}\right|+C

∫dxx(ax+b)2=∫(1b2x−ab2ax+b+−ab(ax+b)2)dx=1b2∫1xdx−ab2∫1ax+bdx−ab∫1(ax+b)2dx=1b(ax+b)−1b2ln⁡∣ax+bx∣+C\begin{align} & \int\dfrac{\mathrm{d}x}{x(ax+b)^2}\\ = & \int\left(\frac{\frac{1}{b^2}}{x}-\frac{\frac{a}{b^2}}{ax+b}+\frac{-\frac{a}{b}}{(ax+b)^2}\right)\mathrm{d}x\\ = & \frac{1}{b^2}\int\frac{1}{x}\mathrm{d}x-\frac{a}{b^2}\int\frac{1}{ax+b}\mathrm{d}x-\frac{a}{b}\int\frac{1}{(ax+b)^2}\mathrm{d}x\\ = & \dfrac{1}{b(ax+b)}-\dfrac{1}{b^2}\ln\left|\dfrac{ax+b}{x}\right|+C \end{align}

（二）含有 ax+b\sqrt{ax+b} 的积分

  1. ∫ax+bdx=23a(ax+b)3+C\displaystyle\int\sqrt{ax+b}\mathrm{d}x=\dfrac{2}{3a}\sqrt{(ax+b)^3}+C

∫ax+bdx=1a∫ax+bd(ax+b)=23a(ax+b)3+C\begin{align} & \int\sqrt{ax+b}\mathrm{d}x\\ = & \frac{1}{a}\int\sqrt{ax+b}\mathrm{d}(ax+b)\\ = & \frac{2}{3a}\sqrt{(ax+b)^3}+C \end{align}

  1. ∫xax+bdx=215a2(3ax−2b)(ax+b)3+C\displaystyle\int x\sqrt{ax+b}\mathrm{d}x=\dfrac{2}{15a^2}(3ax-2b)\sqrt{(ax+b)^3}+C

∫xax+bdx=∫[(1a(ax+b)−ba)ax+b]dx=1a∫(ax+b)32dx−ba∫ax+bdx=25a2(ax+b)52−b3a2(ax+b)32+C=215a2(3ax−2b)(ax+b)3+C\begin{align} & \int x\sqrt{ax+b}\mathrm{d}x\\ = & \int\left[\left(\frac{1}{a}(ax+b)-\frac{b}{a}\right)\sqrt{ax+b}\right]\mathrm{d}x\\ = & \frac{1}{a}\int(ax+b)^{\frac{3}{2}}\mathrm{d}x-\frac{b}{a}\int\sqrt{ax+b}\mathrm{d}x\\ = & \frac{2}{5a^2}(ax+b)^{\frac{5}{2}}-\frac{b}{3a^2}(ax+b)^{\frac{3}{2}}+C\\ = & \dfrac{2}{15a^2}(3ax-2b)\sqrt{(ax+b)^3}+C \end{align}

  1. ∫x2ax+bdx=2105a3(15a2x2−12abx+8b2)(ax+b)3+C\displaystyle\int x^2\sqrt{ax+b}\mathrm{d}x=\dfrac{2}{105a^3}(15a^2x^2-12abx+8b^2)\sqrt{(ax+b)^3}+C

∫x2ax+bdx=∫[1a2(ax+b)2−2ba2(ax+b)+b2a2]ax+bdx=1a2∫(ax+b)52dx−2ba2∫(ax+b)32dx+b2a2∫(ax+b)12dx=27a3(ax+b)72−4b5a3(ax+b)52+2b23a3(ax+b)32=2105a3(15a2x2−12abx+8b2)(ax+b)3+C\begin{align} & \int x^2\sqrt{ax+b}\mathrm{d}x\\ = & \int\left[\frac{1}{a^2}(ax+b)^2-\frac{2b}{a^2}(ax+b)+\frac{b^2}{a^2}\right]\sqrt{ax+b}\mathrm{d}x\\ = & \frac{1}{a^2}\int(ax+b)^{\frac{5}{2}}\mathrm{d}x-\frac{2b}{a^2}\int(ax+b)^{\frac{3}{2}}\mathrm{d}x+\frac{b^2}{a^2}\int(ax+b)^{\frac{1}{2}}\mathrm{d}x\\ = & \frac{2}{7a^3}(ax+b)^{\frac{7}{2}}-\frac{4b}{5a^3}(ax+b)^{\frac{5}{2}}+\frac{2b^2}{3a^3}(ax+b)^{\frac{3}{2}}\\ = & \dfrac{2}{105a^3}(15a^2x^2-12abx+8b^2)\sqrt{(ax+b)^3}+C \end{align}

  1. ∫xax+bdx=23a2(ax−2b)ax+b+C\displaystyle\int\dfrac{x}{\sqrt{ax+b}}\mathrm{d}x=\dfrac{2}{3a^2}(ax-2b)\sqrt{ax+b}+C

∫xax+bdx=∫1a(ax+b)−baax+b=1a∫ax+bdx−ba∫1ax+bdx=23a2(ax+b)32−2ba2(ax+b)12+C=23a2(ax−2b)ax+b+C\begin{align} & \int\dfrac{x}{\sqrt{ax+b}}\mathrm{d}x\\ = & \int\frac{\frac{1}{a}(ax+b)-\frac{b}{a}}{\sqrt{ax+b}}\\ = & \frac{1}{a}\int\sqrt{ax+b}\mathrm{d}x-\frac{b}{a}\int\frac{1}{\sqrt{ax+b}}\mathrm{d}x\\ = & \frac{2}{3a^2}(ax+b)^{\frac{3}{2}}-\frac{2b}{a^2}(ax+b)^{\frac{1}{2}}+C\\ = & \dfrac{2}{3a^2}(ax-2b)\sqrt{ax+b}+C \end{align}

  1. ∫x2ax+bdx=215a3(3a2x2−4abx+8b2)ax+b+C\displaystyle\int\dfrac{x^2}{\sqrt{ax+b}}\mathrm{d}x=\dfrac{2}{15a^3}(3a^2x^2-4abx+8b^2)\sqrt{ax+b}+C

∫x2ax+bdx=∫1a2(ax+b)2−2ba2(ax+b)+b2a2ax+bdx=1a2∫(ax+b)32dx−2ba2∫(ax+b)12dx+b2a2∫(ax+b)−12dx=25a3(ax+b)52−4b3a3(ax+b)32+2b2a3(ax+b)12+C=215a3(3a2x2−4abx+8b2)ax+b+C\begin{align} & \int\dfrac{x^2}{\sqrt{ax+b}}\mathrm{d}x\\ = & \int\frac{\frac{1}{a^2}(ax+b)^2-\frac{2b}{a^2}(ax+b)+\frac{b^2}{a^2}}{\sqrt{ax+b}}\mathrm{d}x\\ = & \frac{1}{a^2}\int(ax+b)^{\frac{3}{2}}\mathrm{d}x-\frac{2b}{a^2}\int(ax+b)^{\frac{1}{2}}\mathrm{d}x+\frac{b^2}{a^2}\int(ax+b)^{-\frac{1}{2}}\mathrm{d}x\\ = & \frac{2}{5a^3}(ax+b)^{\frac{5}{2}}-\frac{4b}{3a^3}(ax+b)^{\frac{3}{2}}+\frac{2b^2}{a^3}(ax+b)^{\frac{1}{2}}+C\\ = & \dfrac{2}{15a^3}(3a^2x^2-4abx+8b^2)\sqrt{ax+b}+C \end{align}

  1. ∫dxxax+b={1bln⁡∣ax+b−bax+b+b∣+C(b>0)2−barctan⁡ax+b−b+C(b<0)\displaystyle\int\dfrac{\mathrm{d}x}{x\sqrt{ax+b}}=\begin{cases}\dfrac{1}{\sqrt{b}}\ln\left|\dfrac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\right|+C&(b>0)\\\dfrac{2}{\sqrt{-b}}\arctan\sqrt{\dfrac{ax+b}{-b}}+C&(b<0)\end{cases}

令 u=ax+bu=\sqrt{ax+b}、则 dx=2uadu\mathrm{d}x=\dfrac{2u}{a}\mathrm{d}u

当 b>0b>0 时

∫dxxax+b=2∫1u2−bdu=2∫1(u−b)(u+b)du=1b∫(1u−b−1u+b)du=1bln⁡∣u−bu+b∣+C==1bln⁡∣ax+b−bax+b+b∣+C\begin{align} \int\dfrac{\mathrm{d}x}{x\sqrt{ax+b}} & = 2\int\frac{1}{u^2-b}\mathrm{d}u\\ & = 2\int\frac{1}{(u-\sqrt{b})(u+\sqrt{b})}\mathrm{d}u\\ & = \frac{1}{\sqrt{b}}\int\left(\frac{1}{u-\sqrt{b}}-\frac{1}{u+\sqrt{b}}\right)\mathrm{d}u\\ & = \frac{1}{\sqrt{b}}\ln\left|\frac{u-\sqrt{b}}{u+\sqrt{b}}\right|+C\\ & = =\dfrac{1}{\sqrt{b}}\ln\left|\dfrac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\right|+C \end{align}

当 b<0b<0 时

∫dxxax+b=2∫1u2+(−b)du=2−b∫11+(u−b)du=2−barctan⁡u−b+C=2−barctan⁡ax+b−b+C\begin{align} \int\dfrac{\mathrm{d}x}{x\sqrt{ax+b}} & = 2\int\frac{1}{u^2+(-b)}\mathrm{d}u\\ & = \frac{2}{-b}\int\frac{1}{1+\left(\frac{u}{\sqrt{-b}}\right)}\mathrm{d}u\\ & = \frac{2}{\sqrt{-b}}\arctan\frac{u}{\sqrt{-b}}+C\\ & = \dfrac{2}{\sqrt{-b}}\arctan\sqrt{\dfrac{ax+b}{-b}}+C \end{align}

  1. ∫dxx2ax+b=−ax+bbx−a2b∫dxxax+b\displaystyle\int\dfrac{\mathrm{d}x}{x^2\sqrt{ax+b}}=-\dfrac{\sqrt{ax+b}}{bx}-\dfrac{a}{2b}\displaystyle\int\dfrac{\mathrm{d}x}{x\sqrt{ax+b}}

∫dxx2ax+b=∫(−abxax+b+1bax+bx2)dx=−ab∫1xax+bdx−1b∫ax+bxd(1x)=−ab∫1xax+bdx−ax+bbx+a2b∫1xax+bdx=−ax+bbx−a2b∫dxxax+b\begin{align} & \int\dfrac{\mathrm{d}x}{x^2\sqrt{ax+b}}\\ = & \int\left(\frac{-\frac{a}{b}}{x\sqrt{ax+b}}+\frac{\frac{1}{b}\sqrt{ax+b}}{x^2}\right)\mathrm{d}x\\ = & -\frac{a}{b}\int\frac{1}{x\sqrt{ax+b}}\mathrm{d}x-\frac{1}{b}\int\frac{\sqrt{ax+b}}{x}\mathrm{d}\left(\dfrac{1}{x}\right)\\ = & -\frac{a}{b}\int\frac{1}{x\sqrt{ax+b}}\mathrm{d}x-\frac{\sqrt{ax+b}}{bx}+\frac{a}{2b}\int\frac{1}{x\sqrt{ax+b}}\mathrm{d}x\\ = & -\dfrac{\sqrt{ax+b}}{bx}-\dfrac{a}{2b}\displaystyle\int\dfrac{\mathrm{d}x}{x\sqrt{ax+b}} \end{align}

  1. ∫ax+bxdx=2ax+b+b∫dxxax+b\displaystyle\int\dfrac{\sqrt{ax+b}}{x}\mathrm{d}x=2\sqrt{ax+b}+b\displaystyle\int\dfrac{\mathrm{d}x}{x\sqrt{ax+b}}

\begin{align} \end{align}

  1. ∫ax+bx2dx=−ax+bx+a2∫dxxax+b\displaystyle\int\dfrac{\sqrt{ax+b}}{x^2}\mathrm{d}x=-\dfrac{\sqrt{ax+b}}{x}+\dfrac{a}{2}\displaystyle\int\dfrac{\mathrm{d}x}{x\sqrt{ax+b}}

\begin{align} \end{align}

（三）含有 x2±a2x^2 \pm a^2 的积分

  1. ∫dxx2+a2=1aarctan⁡xa+C\displaystyle\int\dfrac{\mathrm{d}x}{x^2+a^2}=\dfrac{1}{a}\arctan\dfrac{x}{a}+C

  2. ∫dx(x2+a2)n=x2(n−1)a2(x2+a2)n−1+2n−32(n−1)a2∫dx(x2+a2)n−1\displaystyle\int\dfrac{\mathrm{d}x}{(x^2+a^2)^n}=\dfrac{x}{2(n-1)a^2(x^2+a^2)^{n-1}}+\dfrac{2n-3}{2(n-1)a^2}\displaystyle\int\dfrac{\mathrm{d}x}{(x^2+a^2)^{n-1}}

  3. ∫dxx2−a2=12aln⁡∣x−ax+a∣+C\displaystyle\int\dfrac{\mathrm{d}x}{x^2-a^2}=\dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right|+C

（四）含有 ax2+bax^2+b (a>0)(a>0) 的积分

  1. ∫dxax2+b={1abarctan⁡abx+C(b>0)12−abln⁡∣ax−−bax+−b∣+C(b<0)\displaystyle\int\dfrac{\mathrm{d}x}{ax^2+b}=\begin{cases}\dfrac{1}{\sqrt{ab}}\arctan\sqrt{\dfrac{a}{b}}x+C&(b>0)\\\dfrac{1}{2\sqrt{-ab}}\ln\left|\dfrac{\sqrt{a}x-\sqrt{-b}}{\sqrt{a}x+\sqrt{-b}}\right|+C&(b<0)\end{cases}

  2. ∫xax2+bdx=12aln⁡∣ax2+b∣+C\displaystyle\int\dfrac{x}{ax^2+b}\mathrm{d}x=\dfrac{1}{2a}\ln|ax^2+b|+C

  3. ∫x2ax2+bdx=xa−ba∫dxax2+b\displaystyle\int\dfrac{x^2}{ax^2+b}\mathrm{d}x=\dfrac{x}{a}-\dfrac{b}{a}\displaystyle\int\dfrac{\mathrm{d}x}{ax^2+b}

  4. ∫dxx(ax2+b)=12bln⁡∣x2ax2+b∣+C\displaystyle\int\dfrac{\mathrm{d}x}{x(ax^2+b)}=\dfrac{1}{2b}\ln\left|\dfrac{x^2}{ax^2+b}\right|+C

  5. ∫dxx2(ax2+b)=−1bx−ab∫dxax2+b\displaystyle\int\dfrac{\mathrm{d}x}{x^2(ax^2+b)}=-\dfrac{1}{bx}-\dfrac{a}{b}\displaystyle\int\dfrac{\mathrm{d}x}{ax^2+b}

  6. ∫dxx3(ax2+b)=a2b2ln⁡∣ax2+bx2∣−12bx2+C\displaystyle\int\dfrac{\mathrm{d}x}{x^3(ax^2+b)}=\dfrac{a}{2b^2}\ln\left|\dfrac{ax^2+b}{x^2}\right|-\dfrac{1}{2bx^2}+C

  7. ∫dx(ax2+b)2=x2b(ax2+b)+12b∫dxax2+b\displaystyle\int\dfrac{\mathrm{d}x}{(ax^2+b)^2}=\dfrac{x}{2b(ax^2+b)}+\dfrac{1}{2b}\displaystyle\int\dfrac{\mathrm{d}x}{ax^2+b}

（五）含有 ax2+bx+cax^2+bx+c (a>0)(a>0) 的积分

  1. ∫dxax2+bx+c={24ac−b2arctan⁡2ax+b4ac−b2+C(b2<4ac)1b2−4acln⁡∣2ax+b−b2−4ac2ax+b+b2−4ac∣+C(b2>4ac)\displaystyle\int\dfrac{\mathrm{d}x}{ax^2+bx+c}=\begin{cases}\dfrac{2}{\sqrt{4ac-b^2}}\arctan\dfrac{2ax+b}{\sqrt{4ac-b^2}}+C\quad(b^2<4ac)\\ \dfrac{1}{\sqrt{b^2-4ac}}\ln\left|\dfrac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right|+C\quad(b^2>4ac)\end{cases}

  2. ∫xax2+bx+cdx=12aln⁡∣ax2+bx+c∣−b2a∫dxax2+bx+c\displaystyle\int\dfrac{x}{ax^2+bx+c}\mathrm{d}x=\dfrac{1}{2a}\ln|ax^2+bx+c|-\dfrac{b}{2a}\displaystyle\int\dfrac{\mathrm{d}x}{ax^2+bx+c}

（六）含有 x2+a2\sqrt{x^2+a^2} (a>0)(a>0) 的积分

  1. ∫dxx2+a2=arsh⁡xa+C1=ln⁡(x+x2+a2)+C\displaystyle\int\dfrac{\mathrm{d}x}{\sqrt{x^2+a^2}}=\operatorname{arsh}\dfrac{x}{a}+C_1=\ln(x+\sqrt{x^2+a^2})+C

  2. ∫dx(x2+a2)3=xa2x2+a2+C\displaystyle\int\dfrac{\mathrm{d}x}{\sqrt{(x^2+a^2)^3}}=\dfrac{x}{a^2\sqrt{x^2+a^2}}+C

  3. ∫xx2+a2dx=x2+a2+C\displaystyle\int\dfrac{x}{\sqrt{x^2+a^2}}\mathrm{d}x=\sqrt{x^2+a^2}+C

  4. ∫x(x2+a2)3dx=−1x2+a2+C\displaystyle\int\dfrac{x}{\sqrt{(x^2+a^2)^3}}\mathrm{d}x=-\dfrac{1}{\sqrt{x^2+a^2}}+C

  5. ∫x2x2+a2dx=x2x2+a2−a22ln⁡(x+x2+a2)+C\displaystyle\int\dfrac{x^2}{\sqrt{x^2+a^2}}\mathrm{d}x=\dfrac{x}{2}\sqrt{x^2+a^2}-\dfrac{a^2}{2}\ln(x+\sqrt{x^2+a^2})+C

  6. ∫x2(x2+a2)3dx=−xx2+a2+ln⁡(x+x2+a2)+C\displaystyle\int\dfrac{x^2}{\sqrt{(x^2+a^2)^3}}\mathrm{d}x=-\dfrac{x}{\sqrt{x^2+a^2}}+\ln(x+\sqrt{x^2+a^2})+C

  7. ∫dxxx2+a2=1aln⁡x2+a2−a∣x∣+C\displaystyle\int\dfrac{\mathrm{d}x}{x\sqrt{x^2+a^2}}=\dfrac{1}{a}\ln\dfrac{\sqrt{x^2+a^2}-a}{|x|}+C

  8. ∫dxx2x2+a2=−x2+a2a2x+C\displaystyle\int\dfrac{\mathrm{d}x}{x^2\sqrt{x^2+a^2}}=-\dfrac{\sqrt{x^2+a^2}}{a^2x}+C

  9. ∫x2+a2dx=x2x2+a2+a22ln⁡(x+x2+a2)+C\displaystyle\int\sqrt{x^2+a^2}\mathrm{d}x=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}\ln(x+\sqrt{x^2+a^2})+C

  10. ∫(x2+a2)3dx=x8(2x2+5a2)x2+a2+38a4ln⁡(x+x2+a2)+C\displaystyle\int\sqrt{(x^2+a^2)^3}\mathrm{d}x=\dfrac{x}{8}(2x^2+5a^2)\sqrt{x^2+a^2}+\dfrac{3}{8}a^4\ln(x+\sqrt{x^2+a^2})+C

  11. ∫xx2+a2dx=13(x2+a2)3+C\displaystyle\int x\sqrt{x^2+a^2}\mathrm{d}x=\dfrac{1}{3}\sqrt{(x^2+a^2)^3}+C

  12. ∫x2x2+a2dx=x8(2x2+a2)x2+a2−a48ln⁡(x+x2+a2)+C\displaystyle\int x^2\sqrt{x^2+a^2}\mathrm{d}x=\dfrac{x}{8}(2x^2+a^2)\sqrt{x^2+a^2}-\dfrac{a^4}{8}\ln(x+\sqrt{x^2+a^2})+C

  13. ∫x2+a2xdx=x2+a2+aln⁡x2+a2−a∣x∣+C\displaystyle\int\dfrac{\sqrt{x^2+a^2}}{x}\mathrm{d}x=\sqrt{x^2+a^2}+a\ln\dfrac{\sqrt{x^2+a^2}-a}{|x|}+C

  14. ∫x2+a2x2dx=−x2+a2x+ln⁡(x+x2+a2)+C\displaystyle\int\dfrac{\sqrt{x^2+a^2}}{x^2}\mathrm{d}x=-\dfrac{\sqrt{x^2+a^2}}{x}+\ln(x+\sqrt{x^2+a^2})+C

（七）含有 x2−a2\sqrt{x^2-a^2} (a>0)(a>0) 的积分

  1. ∫dxx2−a2=x∣x∣arch∣x∣a+C1=ln⁡∣x+x2−a2∣+C\displaystyle\int\dfrac{\mathrm{d}x}{\sqrt{x^2-a^2}}=\dfrac{x}{|x|}\mathrm{arch} \dfrac{|x|}{a}+C_1=\ln|x+\sqrt{x^2-a^2}|+C

  2. ∫dx(x2−a2)3=−xa2x2−a2+C\displaystyle\int\dfrac{\mathrm{d}x}{\sqrt{(x^2-a^2)^3}}=-\dfrac{x}{a^2\sqrt{x^2-a^2}}+C

  3. ∫xx2−a2dx=x2−a2+C\displaystyle\int\dfrac{x}{\sqrt{x^2-a^2}}\mathrm{d}x=\sqrt{x^2-a^2}+C

  4. ∫x(x2−a2)3dx=−1x2−a2+C\displaystyle\int\dfrac{x}{\sqrt{(x^2-a^2)^3}}\mathrm{d}x=-\dfrac{1}{\sqrt{x^2-a^2}}+C

  5. ∫x2x2−a2dx=x2x2−a2+a22ln⁡∣x+x2−a2∣+C\displaystyle\int\dfrac{x^2}{\sqrt{x^2-a^2}}\mathrm{d}x=\dfrac{x}{2}\sqrt{x^2-a^2}+\dfrac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|+C

  6. ∫x2(x2−a2)3dx=−xx2−a2+ln⁡∣x+x2−a2∣+C\displaystyle\int\dfrac{x^2}{\sqrt{(x^2-a^2)^3}}\mathrm{d}x=-\dfrac{x}{\sqrt{x^2-a^2}}+\ln|x+\sqrt{x^2-a^2}|+C

  7. ∫dxxx2−a2=1aarccos⁡a∣x∣+C\displaystyle\int\dfrac{\mathrm{d}x}{x\sqrt{x^2-a^2}}=\dfrac{1}{a}\arccos\dfrac{a}{|x|}+C

  8. ∫dxx2x2−a2=x2−a2a2x+C\displaystyle\int\dfrac{\mathrm{d}x}{x^2\sqrt{x^2-a^2}}=\dfrac{\sqrt{x^2-a^2}}{a^2x}+C

  9. ∫x2−a2dx=x2x2−a2−a22ln⁡∣x+x2−a2∣+C\displaystyle\int\sqrt{x^2-a^2}\mathrm{d}x=\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|+C

  10. ∫(x2−a2)3dx=x8(2x2−5a2)x2−a2+38a4ln⁡∣x+x2−a2∣+C\displaystyle\int\sqrt{(x^2-a^2)^3}\mathrm{d}x=\dfrac{x}{8}(2x^2-5a^2)\sqrt{x^2-a^2}+\dfrac{3}{8}a^4\ln|x+\sqrt{x^2-a^2}|+C

  11. ∫xx2−a2dx=13(x2−a2)3+C\displaystyle\int x\sqrt{x^2-a^2}\mathrm{d}x=\dfrac{1}{3}\sqrt{(x^2-a^2)^3}+C

  12. ∫x2x2−a2dx=x8(2x2−a2)x2−a2−a48ln⁡∣x+x2−a2∣+C\displaystyle\int x^2\sqrt{x^2-a^2}\mathrm{d}x=\dfrac{x}{8}(2x^2-a^2)\sqrt{x^2-a^2}-\dfrac{a^4}{8}\ln|x+\sqrt{x^2-a^2}|+C

  13. ∫x2−a2xdx=x2−a2−aarccos⁡a∣x∣+C\displaystyle\int\dfrac{\sqrt{x^2-a^2}}{x}\mathrm{d}x=\sqrt{x^2-a^2}-a\arccos\dfrac{a}{|x|}+C

  14. ∫x2−a2x2dx=−x2−a2x+ln⁡∣x+x2−a2∣+C\displaystyle\int\dfrac{\sqrt{x^2-a^2}}{x^2}\mathrm{d}x=-\dfrac{\sqrt{x^2-a^2}}{x}+\ln|x+\sqrt{x^2-a^2}|+C

（八）含有 a2−x2\sqrt{a^2-x^2} (a>0)(a>0) 的积分

  1. ∫dxa2−x2=arcsin⁡xa+C\displaystyle\int\dfrac{\mathrm{d}x}{\sqrt{a^2-x^2}}=\arcsin\dfrac{x}{a}+C

  2. ∫dx(a2−x2)3=xa2a2−x2+C\displaystyle\int\dfrac{\mathrm{d}x}{\sqrt{(a^2-x^2)^3}}=\dfrac{x}{a^2\sqrt{a^2-x^2}}+C

  3. ∫xa2−x2dx=−a2−x2+C\displaystyle\int\dfrac{x}{\sqrt{a^2-x^2}}\mathrm{d}x=-\sqrt{a^2-x^2}+C

  4. ∫x2a2−x2dx=−x2a2−x2+a22arcsin⁡xa+C\displaystyle\int\dfrac{x^2}{\sqrt{a^2-x^2}}\mathrm{d}x=-\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\arcsin\dfrac{x}{a}+C

  5. ∫x2(a2−x2)3dx=xa2−x2−arcsin⁡xa+C\displaystyle\int\dfrac{x^2}{\sqrt{(a^2-x^2)^3}}\mathrm{d}x=\dfrac{x}{\sqrt{a^2-x^2}}-\arcsin\dfrac{x}{a}+C

  6. ∫dxxa2−x2=1aln⁡∣a−a2−x2∣x∣∣+C\displaystyle\int\dfrac{\mathrm{d}x}{x\sqrt{a^2-x^2}}=\dfrac{1}{a}\ln\left|\dfrac{a-\sqrt{a^2-x^2}}{|x|}\right|+C

  7. ∫dxx2a2−x2=−a2−x2a2x+C\displaystyle\int\dfrac{\mathrm{d}x}{x^2\sqrt{a^2-x^2}}=-\dfrac{\sqrt{a^2-x^2}}{a^2x}+C

  8. ∫a2−x2dx=x2a2−x2+a22arcsin⁡xa+C\displaystyle\int\sqrt{a^2-x^2}\mathrm{d}x=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\arcsin\dfrac{x}{a}+C

  9. ∫(a2−x2)3dx=x8(5a2−2x2)a2−x2+38a4arcsin⁡xa+C\displaystyle\int\sqrt{(a^2-x^2)^3}\mathrm{d}x=\dfrac{x}{8}(5a^2-2x^2)\sqrt{a^2-x^2}+\dfrac{3}{8}a^4\arcsin\dfrac{x}{a}+C

  10. ∫xa2−x2dx=−13(a2−x2)3+C\displaystyle\int x\sqrt{a^2-x^2}\mathrm{d}x=-\dfrac{1}{3}\sqrt{(a^2-x^2)^3}+C

  11. ∫x2a2−x2dx=x8(2x2−a2)a2−x2+a48arcsin⁡xa+C\displaystyle\int x^2\sqrt{a^2-x^2}\mathrm{d}x=\dfrac{x}{8}(2x^2-a^2)\sqrt{a^2-x^2}+\dfrac{a^4}{8}\arcsin\dfrac{x}{a}+C

  12. ∫a2−x2xdx=a2−x2+aln⁡∣a−a2−x2x∣+C\displaystyle\int\dfrac{\sqrt{a^2-x^2}}{x}\mathrm{d}x=\sqrt{a^2-x^2}+a\ln\left|\dfrac{a-\sqrt{a^2-x^2}}{x}\right|+C

  13. ∫a2−x2x2dx=−a2−x2x−arcsin⁡xa+C\displaystyle\int\dfrac{\sqrt{a^2-x^2}}{x^2}\mathrm{d}x=-\dfrac{\sqrt{a^2-x^2}}{x}-\arcsin\dfrac{x}{a}+C

含有 ( \sqrt{ax^2+bx+c} ) (( a>0 )) 的积分

  1. ∫dxax2+bx+c=1aln⁡∣2ax+b+2aax2+bx+c∣+C\displaystyle\int\dfrac{\mathrm{d}x}{\sqrt{ax^2+bx+c}}=\dfrac{1}{\sqrt{a}}\ln\left|2ax+b+2\sqrt{a}\sqrt{ax^2+bx+c}\right|+C

  2. ∫ax2+bx+cdx=2ax+b4aax2+bx+c+4ac−b28a3ln⁡∣2ax+b+2aax2+bx+c∣+C\displaystyle\int\sqrt{ax^2+bx+c}\mathrm{d}x=\dfrac{2ax+b}{4a}\sqrt{ax^2+bx+c}+\dfrac{4ac-b^2}{8\sqrt{a^3}}\ln\left|2ax+b+2\sqrt{a}\sqrt{ax^2+bx+c}\right|+C

  3. ∫xax2+bx+cdx=1aax2+bx+c−b2a3ln⁡∣2ax+b+2aax2+bx+c∣+C\displaystyle\int\dfrac{x}{\sqrt{ax^2+bx+c}}\mathrm{d}x=\dfrac{1}{a}\sqrt{ax^2+bx+c}-\dfrac{b}{2\sqrt{a^3}}\ln\left|2ax+b+2\sqrt{a}\sqrt{ax^2+bx+c}\right|+C

  4. ∫dxc+bx−ax2=1aarcsin⁡2ax−bb2+4ac+C\displaystyle\int\dfrac{\mathrm{d}x}{\sqrt{c+bx-ax^2}}=\dfrac{1}{\sqrt{a}}\arcsin\dfrac{2ax-b}{\sqrt{b^2+4ac}}+C

  5. ∫c+bx−ax2dx=2ax−b4ac+bx−ax2+b2+4ac8a3arcsin⁡2ax−bb2+4ac+C\displaystyle\int\sqrt{c+bx-ax^2}\mathrm{d}x=\dfrac{2ax-b}{4a}\sqrt{c+bx-ax^2}+\dfrac{b^2+4ac}{8\sqrt{a^3}}\arcsin\dfrac{2ax-b}{\sqrt{b^2+4ac}}+C

  6. ∫xc+bx−ax2dx=−1ac+bx−ax2+b2a3arcsin⁡2ax−bb2+4ac+C\displaystyle\int\dfrac{x}{\sqrt{c+bx-ax^2}}\mathrm{d}x=-\dfrac{1}{a}\sqrt{c+bx-ax^2}+\dfrac{b}{2\sqrt{a^3}}\arcsin\dfrac{2ax-b}{\sqrt{b^2+4ac}}+C

含有 ( \sqrt{\dfrac{x-a}{x-b}} ) 或 ( \sqrt{(x-a)(b-x)} ) 的积分

  1. ∫x−ax−bdx=(x−b)x−ax−b+(b−a)ln⁡(∣x−a∣+∣x−b∣)+C\displaystyle\int\sqrt{\dfrac{x-a}{x-b}}\mathrm{d}x=(x-b)\sqrt{\dfrac{x-a}{x-b}}+(b-a)\ln(\sqrt{|x-a|}+\sqrt{|x-b|})+C

  2. ∫x−ab−xdx=(x−b)x−ab−x+(b−a)arcsin⁡x−ab−a+C\displaystyle\int\sqrt{\dfrac{x-a}{b-x}}\mathrm{d}x=(x-b)\sqrt{\dfrac{x-a}{b-x}}+(b-a)\arcsin\sqrt{\dfrac{x-a}{b-a}}+C

  3. ∫dx(x−a)(b−x)=2arcsin⁡x−ab−a+C(a<b)\displaystyle\int\dfrac{\mathrm{d}x}{\sqrt{(x-a)(b-x)}}=2\arcsin\sqrt{\dfrac{x-a}{b-a}}+C\quad(a<b)

  4. ∫(x−a)(b−x)dx=2x−a−b4(x−a)(b−x)+(b−a)24arcsin⁡x−ab−a+C(a<b)\displaystyle\int\sqrt{(x-a)(b-x)}\mathrm{d}x=\dfrac{2x-a-b}{4}\sqrt{(x-a)(b-x)}+\dfrac{(b-a)^2}{4}\arcsin\sqrt{\dfrac{x-a}{b-a}}+C\quad(a<b)

含有三角函数的积分

  1. ∫sin⁡xdx=−cos⁡x+C\displaystyle\int\sin x\mathrm{d}x=-\cos x+C

  2. ∫cos⁡xdx=sin⁡x+C\displaystyle\int\cos x\mathrm{d}x=\sin x+C

  3. ∫tan⁡xdx=−ln⁡∣cos⁡x∣+C\displaystyle\int\tan x\mathrm{d}x=-\ln|\cos x|+C

  4. ∫cot⁡xdx=ln⁡∣sin⁡x∣+C\displaystyle\int\cot x\mathrm{d}x=\ln|\sin x|+C

  5. ∫sec⁡xdx=ln⁡∣tan⁡(π4+x2)∣+C=ln⁡∣sec⁡x+tan⁡x∣+C\displaystyle\int\sec x\mathrm{d}x=\ln\left|\tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\right|+C=\ln|\sec x+\tan x|+C

  6. ∫csc⁡xdx=ln⁡∣tan⁡x2∣+C=ln⁡∣csc⁡x−cot⁡x∣+C\displaystyle\int\csc x\mathrm{d}x=\ln\left|\tan\dfrac{x}{2}\right|+C=\ln|\csc x-\cot x|+C

  7. ∫sec⁡2xdx=tan⁡x+C\displaystyle\int\sec^2 x\mathrm{d}x=\tan x+C

  8. ∫csc⁡2xdx=−cot⁡x+C\displaystyle\int\csc^2 x\mathrm{d}x=-\cot x+C

  9. ∫sec⁡xtan⁡xdx=sec⁡x+C\displaystyle\int\sec x\tan x\mathrm{d}x=\sec x+C

  10. ∫csc⁡xcot⁡xdx=−csc⁡x+C\displaystyle\int\csc x\cot x\mathrm{d}x=-\csc x+C

  11. ∫sin⁡2xdx=x2−14sin⁡2x+C\displaystyle\int\sin^2 x\mathrm{d}x=\dfrac{x}{2}-\dfrac{1}{4}\sin 2x+C

  12. ∫cos⁡2xdx=x2+14sin⁡2x+C\displaystyle\int\cos^2 x\mathrm{d}x=\dfrac{x}{2}+\dfrac{1}{4}\sin 2x+C

  13. ∫sin⁡nxdx=−1nsin⁡n−1xcos⁡x+n−1n∫sin⁡n−2xdx\displaystyle\int\sin^n x\mathrm{d}x=-\dfrac{1}{n}\sin^{n-1}x\cos x+\dfrac{n-1}{n}\displaystyle\int\sin^{n-2}x\mathrm{d}x

  14. ∫cos⁡nxdx=1ncos⁡n−1xsin⁡x+n−1n∫cos⁡n−2xdx\displaystyle\int\cos^n x\mathrm{d}x=\dfrac{1}{n}\cos^{n-1}x\sin x+\dfrac{n-1}{n}\displaystyle\int\cos^{n-2}x\mathrm{d}x

  15. ∫dxsin⁡nx=−1n−1⋅cos⁡xsin⁡n−1x+n−2n−1∫dxsin⁡n−2x\displaystyle\int\dfrac{\mathrm{d}x}{\sin^n x}=-\dfrac{1}{n-1}\cdot\dfrac{\cos x}{\sin^{n-1}x}+\dfrac{n-2}{n-1}\displaystyle\int\dfrac{\mathrm{d}x}{\sin^{n-2}x}

  16. ∫dxcos⁡nx=1n−1⋅sin⁡xcos⁡n−1x+n−2n−1∫dxcos⁡n−2x\displaystyle\int\dfrac{\mathrm{d}x}{\cos^n x}=\dfrac{1}{n-1}\cdot\dfrac{\sin x}{\cos^{n-1}x}+\dfrac{n-2}{n-1}\displaystyle\int\dfrac{\mathrm{d}x}{\cos^{n-2}x}

  17. ∫cos⁡mxsin⁡nxdx=1m+ncos⁡m−1xsin⁡n+1x+m−1m+n∫cos⁡m−2xsin⁡nxdx=−1m+ncos⁡m+1xsin⁡n−1x+n−1m+n∫cos⁡mxsin⁡n−2xdx\displaystyle\int\cos^m x\sin^n x\mathrm{d}x=\dfrac{1}{m+n}\cos^{m-1}x\sin^{n+1}x+\dfrac{m-1}{m+n}\displaystyle\int\cos^{m-2}x\sin^n x\mathrm{d}x=-\dfrac{1}{m+n}\cos^{m+1}x\sin^{n-1}x+\dfrac{n-1}{m+n}\displaystyle\int\cos^m x\sin^{n-2}x\mathrm{d}x

  18. ∫sin⁡axcos⁡bxdx=−12(a+b)cos⁡(a+b)x−12(a−b)cos⁡(a−b)x+C\displaystyle\int\sin ax\cos bx\mathrm{d}x=-\dfrac{1}{2(a+b)}\cos(a+b)x-\dfrac{1}{2(a-b)}\cos(a-b)x+C

  19. ∫sin⁡axsin⁡bxdx=−12(a+b)sin⁡(a+b)x+12(a−b)sin⁡(a−b)x+C\displaystyle\int\sin ax\sin bx\mathrm{d}x=-\dfrac{1}{2(a+b)}\sin(a+b)x+\dfrac{1}{2(a-b)}\sin(a-b)x+C

  20. ∫cos⁡axcos⁡bxdx=12(a+b)sin⁡(a+b)x+12(a−b)sin⁡(a−b)x+C\displaystyle\int\cos ax\cos bx\mathrm{d}x=\dfrac{1}{2(a+b)}\sin(a+b)x+\dfrac{1}{2(a-b)}\sin(a-b)x+C

  21. ∫dxa+bsin⁡x=2a2−b2arctan⁡tan⁡x2a2−b2+C(a2>b2)\displaystyle\int\dfrac{\mathrm{d}x}{a+b\sin x}=\dfrac{2}{\sqrt{a^2-b^2}}\arctan\dfrac{\tan\dfrac{x}{2}}{\sqrt{a^2-b^2}}+C\quad(a^2>b^2)

  22. ∫dxa+bsin⁡x=1b2−a2ln⁡∣tan⁡x2a2−b2+tan⁡x2a2−b2∣+C(a2<b2)\displaystyle\int\dfrac{\mathrm{d}x}{a+b\sin x}=\dfrac{1}{\sqrt{b^2-a^2}}\ln\left|\dfrac{\tan\dfrac{x}{2}}{\sqrt{a^2-b^2}}+\dfrac{\tan\dfrac{x}{2}}{\sqrt{a^2-b^2}}\right|+C\quad(a^2<b^2)

  23. ∫dxa+bcos⁡x=2a+ba+ba−barctan⁡(a−ba+btan⁡x2)+C(a2>b2)\displaystyle\int\dfrac{\mathrm{d}x}{a+b\cos x}=\dfrac{2}{a+b}\sqrt{\dfrac{a+b}{a-b}}\arctan\left(\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{x}{2}\right)+C\quad(a^2>b^2)

  24. ∫dxa+bcos⁡x=1a+ba+bb−aln⁡∣tan⁡x2a2−b2+tan⁡x2a2−b2∣+C(a2<b2)\displaystyle\int\dfrac{\mathrm{d}x}{a+b\cos x}=\dfrac{1}{a+b}\sqrt{\dfrac{a+b}{b-a}}\ln\left|\dfrac{\tan\dfrac{x}{2}}{\sqrt{a^2-b^2}}+\dfrac{\tan\dfrac{x}{2}}{\sqrt{a^2-b^2}}\right|+C\quad(a^2<b^2)

  25. ∫dxa2cos⁡2x+b2sin⁡2x=1abarctan⁡(batan⁡x)+C\displaystyle\int\dfrac{\mathrm{d}x}{a^2\cos^2 x+b^2\sin^2 x}=\dfrac{1}{ab}\arctan\left(\dfrac{b}{a}\tan x\right)+C

  26. ∫dxa2cos⁡2x−b2sin⁡2x=12abln⁡∣btan⁡x+abtan⁡x−a∣+C\displaystyle\int\dfrac{\mathrm{d}x}{a^2\cos^2 x-b^2\sin^2 x}=\dfrac{1}{2ab}\ln\left|\dfrac{b\tan x+a}{b\tan x-a}\right|+C

  27. ∫xsin⁡axdx=1a2sin⁡ax−1axcos⁡ax+C\displaystyle\int x\sin ax\mathrm{d}x=\dfrac{1}{a^2}\sin ax-\dfrac{1}{a}x\cos ax+C

  28. ∫x2sin⁡axdx=−1ax2cos⁡ax+2a2xsin⁡ax+2a3cos⁡ax+C\displaystyle\int x^2\sin ax\mathrm{d}x=-\dfrac{1}{a}x^2\cos ax+\dfrac{2}{a^2}x\sin ax+\dfrac{2}{a^3}\cos ax+C

  29. ∫xcos⁡axdx=1a2cos⁡ax+1axsin⁡ax+C\displaystyle\int x\cos ax\mathrm{d}x=\dfrac{1}{a^2}\cos ax+\dfrac{1}{a}x\sin ax+C

  30. ∫x2cos⁡axdx=1ax2sin⁡ax+2a2xcos⁡ax−2a3sin⁡ax+C\displaystyle\int x^2\cos ax\mathrm{d}x=\dfrac{1}{a}x^2\sin ax+\dfrac{2}{a^2}x\cos ax-\dfrac{2}{a^3}\sin ax+C

含有反三角函数的积分（其中 ( a>0 )）

  1. ∫arcsin⁡xadx=xarcsin⁡xa+a2−x2+C\displaystyle\int\arcsin\dfrac{x}{a}\mathrm{d}x=x\arcsin\dfrac{x}{a}+\sqrt{a^2-x^2}+C

  2. ∫xarcsin⁡xadx=(x22−a24)arcsin⁡xa+x4a2−x2+C\displaystyle\int x\arcsin\dfrac{x}{a}\mathrm{d}x=\left(\dfrac{x^2}{2}-\dfrac{a^2}{4}\right)\arcsin\dfrac{x}{a}+\dfrac{x}{4}\sqrt{a^2-x^2}+C

  3. ∫x2arcsin⁡xadx=x33arcsin⁡xa+19(x2+2a2)a2−x2+C\displaystyle\int x^2\arcsin\dfrac{x}{a}\mathrm{d}x=\dfrac{x^3}{3}\arcsin\dfrac{x}{a}+\dfrac{1}{9}(x^2+2a^2)\sqrt{a^2-x^2}+C

  4. ∫arccos⁡xadx=xarccos⁡xa−a2−x2+C\displaystyle\int\arccos\dfrac{x}{a}\mathrm{d}x=x\arccos\dfrac{x}{a}-\sqrt{a^2-x^2}+C

  5. ∫xarccos⁡xadx=(x22−a24)arccos⁡xa−x4a2−x2+C\displaystyle\int x\arccos\dfrac{x}{a}\mathrm{d}x=\left(\dfrac{x^2}{2}-\dfrac{a^2}{4}\right)\arccos\dfrac{x}{a}-\dfrac{x}{4}\sqrt{a^2-x^2}+C

  6. ∫x2arccos⁡xadx=x33arccos⁡xa−19(x2+2a2)a2−x2+C\displaystyle\int x^2\arccos\dfrac{x}{a}\mathrm{d}x=\dfrac{x^3}{3}\arccos\dfrac{x}{a}-\dfrac{1}{9}(x^2+2a^2)\sqrt{a^2-x^2}+C

  7. ∫arctan⁡xadx=xarctan⁡xa−a2ln⁡(a2+x2)+C\displaystyle\int\arctan\dfrac{x}{a}\mathrm{d}x=x\arctan\dfrac{x}{a}-\dfrac{a}{2}\ln(a^2+x^2)+C

  8. ∫xarctan⁡xadx=12(a2+x2)arctan⁡xa−a2x+C\displaystyle\int x\arctan\dfrac{x}{a}\mathrm{d}x=\dfrac{1}{2}(a^2+x^2)\arctan\dfrac{x}{a}-\dfrac{a}{2}x+C

  9. ∫x2arctan⁡xadx=x33arctan⁡xa−a6x2+a36ln⁡(a2+x2)+C\displaystyle\int x^2\arctan\dfrac{x}{a}\mathrm{d}x=\dfrac{x^3}{3}\arctan\dfrac{x}{a}-\dfrac{a}{6}x^2+\dfrac{a^3}{6}\ln(a^2+x^2)+C

含有指数函数的积分

  1. ∫axdx=1ln⁡aax+C\displaystyle\int a^x\mathrm{d}x=\dfrac{1}{\ln a}a^x+C

  2. ∫eaxdx=1aeax+C\displaystyle\int e^{ax}\mathrm{d}x=\dfrac{1}{a}e^{ax}+C

  3. ∫xeaxdx=1a2(ax−1)eax+C\displaystyle\int xe^{ax}\mathrm{d}x=\dfrac{1}{a^2}(ax-1)e^{ax}+C

  4. ∫xneaxdx=1axneax−na∫xn−1eaxdx\displaystyle\int x^n e^{ax}\mathrm{d}x=\dfrac{1}{a}x^n e^{ax}-\dfrac{n}{a}\displaystyle\int x^{n-1}e^{ax}\mathrm{d}x

  5. ∫xaxdx=xln⁡aax−1(ln⁡a)2ax+C\displaystyle\int x a^x\mathrm{d}x=\dfrac{x}{\ln a}a^x-\dfrac{1}{(\ln a)^2}a^x+C

  6. ∫xnaxdx=1ln⁡axnax−nln⁡a∫xn−1axdx\displaystyle\int x^n a^x\mathrm{d}x=\dfrac{1}{\ln a}x^n a^x-\dfrac{n}{\ln a}\displaystyle\int x^{n-1}a^x\mathrm{d}x

  7. ∫eaxsin⁡bxdx=1a2+b2eax(asin⁡bx−bcos⁡bx)+C\displaystyle\int e^{ax}\sin bx\mathrm{d}x=\dfrac{1}{a^2+b^2}e^{ax}(a\sin bx-b\cos bx)+C

  8. ∫eaxcos⁡bxdx=1a2+b2eax(acos⁡bx+bsin⁡bx)+C\displaystyle\int e^{ax}\cos bx\mathrm{d}x=\dfrac{1}{a^2+b^2}e^{ax}(a\cos bx+b\sin bx)+C

  9. ∫eaxsin⁡nbxdx=1a2+b2n2eaxsin⁡n−1bx(asin⁡bx−nbcos⁡bx)+n(n−1)b2a2+b2n2∫eaxsin⁡n−2bxdx\displaystyle\int e^{ax}\sin^n bx\mathrm{d}x=\dfrac{1}{a^2+b^2n^2}e^{ax}\sin^{n-1}bx(a\sin bx-nb\cos bx)+\dfrac{n(n-1)b^2}{a^2+b^2n^2}\displaystyle\int e^{ax}\sin^{n-2}bx\mathrm{d}x

  10. ∫eaxcos⁡nbxdx=1a2+b2n2eaxcos⁡n−1bx(acos⁡bx+nbsin⁡bx)+n(n−1)b2a2+b2n2∫eaxcos⁡n−2bxdx\displaystyle\int e^{ax}\cos^n bx\mathrm{d}x=\dfrac{1}{a^2+b^2n^2}e^{ax}\cos^{n-1}bx(a\cos bx+nb\sin bx)+\dfrac{n(n-1)b^2}{a^2+b^2n^2}\displaystyle\int e^{ax}\cos^{n-2}bx\mathrm{d}x

含有对数函数的积分

  1. ∫ln⁡xdx=xln⁡x−x+C\displaystyle\int\ln x\mathrm{d}x=x\ln x-x+C

  2. ∫dxxln⁡x=ln⁡∣ln⁡x∣+C\displaystyle\int\dfrac{\mathrm{d}x}{x\ln x}=\ln|\ln x|+C

  3. ∫xnln⁡xdx=1n+1xn+1(ln⁡x−1n+1)+C\displaystyle\int x^n\ln x\mathrm{d}x=\dfrac{1}{n+1}x^{n+1}\left(\ln x-\dfrac{1}{n+1}\right)+C

  4. ∫(ln⁡x)ndx=x(ln⁡x)n−n∫(ln⁡x)n−1dx\displaystyle\int(\ln x)^n\mathrm{d}x=x(\ln x)^n-n\displaystyle\int(\ln x)^{n-1}\mathrm{d}x

  5. ∫xm(ln⁡x)ndx=1m+1xm+1(ln⁡x)n−nm+1∫xm(ln⁡x)n−1dx\displaystyle\int x^m(\ln x)^n\mathrm{d}x=\dfrac{1}{m+1}x^{m+1}(\ln x)^n-\dfrac{n}{m+1}\displaystyle\int x^m(\ln x)^{n-1}\mathrm{d}x

含有双曲函数的积分

  1. ∫sh⁡xdx=ch⁡x+C\displaystyle\int\operatorname{sh} x\mathrm{d}x=\operatorname{ch} x+C

  2. ∫ch⁡xdx=sh⁡x+C\displaystyle\int\operatorname{ch} x\mathrm{d}x=\operatorname{sh} x+C

  3. ∫th⁡xdx=ln⁡ch⁡x+C\displaystyle\int\operatorname{th} x\mathrm{d}x=\ln\operatorname{ch} x+C

  4. ∫sh⁡2xdx=−x2+14sh⁡2x+C\displaystyle\int\operatorname{sh}^2 x\mathrm{d}x=-\dfrac{x}{2}+\dfrac{1}{4}\operatorname{sh} 2x+C

  5. ∫ch⁡2xdx=x2+14sh⁡2x+C\displaystyle\int\operatorname{ch}^2 x\mathrm{d}x=\dfrac{x}{2}+\dfrac{1}{4}\operatorname{sh} 2x+C

（十六）定积分

  1. ∫−ππcos⁡nxdx=∫−ππsin⁡nxdx=0\displaystyle\int_{-\pi}^{\pi}\cos nx\mathrm{d}x=\displaystyle\int_{-\pi}^{\pi}\sin nx\mathrm{d}x=0

  2. ∫−ππcos⁡mxsin⁡nxdx=0\displaystyle\int_{-\pi}^{\pi}\cos mx\sin nx\mathrm{d}x=0

  3. ∫−ππcos⁡mxcos⁡nxdx={0,m≠nπ,m=n\displaystyle\int_{-\pi}^{\pi}\cos mx\cos nx\mathrm{d}x=\begin{cases}0,&m\neq n\\\pi,&m=n\end{cases}

  4. ∫−ππsin⁡mxsin⁡nxdx={0,m≠nπ,m=n\displaystyle\int_{-\pi}^{\pi}\sin mx\sin nx\mathrm{d}x=\begin{cases}0,&m\neq n\\\pi,&m=n\end{cases}

  5. ∫0πsin⁡mxsin⁡nxdx=∫0πcos⁡mxcos⁡nxdx={0,m≠nπ2,m=n\displaystyle\int_0^{\pi}\sin mx\sin nx\mathrm{d}x=\displaystyle\int_0^{\pi}\cos mx\cos nx\mathrm{d}x=\begin{cases}0,&m\neq n\\\dfrac{\pi}{2},&m=n\end{cases}

  6. In=∫0π2sin⁡nxdx=∫0π2cos⁡nxdx,In=n−1nIn−2\displaystyle I_n=\displaystyle\int_0^{\frac{\pi}{2}}\sin^n x\mathrm{d}x=\displaystyle\int_0^{\frac{\pi}{2}}\cos^n x\mathrm{d}x,\quad I_n=\dfrac{n-1}{n}I_{n-2}