求积分

∫xln⁡(x+1+x2)(1+x2)2dx \int \frac{x\ln(x+\sqrt{1+x^2})}{(1+x^2)^2}dx

解：

∫xln⁡(x+1+x2)(1+x2)2dx=∫ln⁡(x+1+x2)2(1+x2)2d(x2)=∫ln⁡(x+1+x2)2(1+x2)2d(x2+1)=−12∫ln⁡(x+1+x2)d(11+x2)=−ln⁡(x+1+x2)⋅11+x2+12∫11+x2⋅1+x1+x21+1+x2dx=−12ln⁡(x+1+x2)1+x2+12∫1(1+x2)32dx \begin{aligned} & \int \frac{x\ln(x+\sqrt{1+x^2})}{(1+x^2)^2}dx \\ = & \int\frac{\ln(x+\sqrt{1+x^2})}{2(1+x^2)^2}d(x^2)\\ = &\int\frac{\ln(x+\sqrt{1+x^2})}{2(1+x^2)^2}d(x^2+1) \\ = & -\frac{1}{2}\int \ln(x+\sqrt{1+x^2})d\left(\frac{1}{1+x^2} \right) \\ = & -\ln(x+\sqrt{1+x^2})\cdot \frac{1}{1+x^2} + \frac{1}{2}\int \frac{1}{1+x^2}\cdot \frac{1+\frac{x}{\sqrt{1+x^2}}}{1+\sqrt{1+x^2}}dx \\ = & -\frac{1}{2}\frac{\ln(x+\sqrt{1+x^2})}{1+x^2} + \frac{1}{2}\int \frac{1}{(1+x^2)^{\frac{3}{2}}}dx \end{aligned}

另解 ∫1(1+x2)32dx\int \frac{1}{(1+x^2)^{\frac{3}{2}}}dx

∫1(1+x2)32dx=∫1(1+tan⁡2θ)32dtan⁡θ=x=tan⁡θ∫1sec⁡3θsec⁡2θdθ=∫cos⁡θdθ=sin⁡θ \begin{aligned} & \int \frac{1}{(1+x^2)^{\frac{3}{2}}}dx \\ =& \int \frac{1}{(1+\tan^2\theta)^{\frac{3}{2}}} d\tan\theta \\ \overset{x = \tan\theta}{=} & \int \frac{1}{\sec^3\theta}\sec^2 \theta d\theta = \int \cos\theta d\theta \\ =& \sin\theta \end{aligned}

有

x2=sin⁡2θ1−sin⁡2θ⇔x2−x2sin⁡2θ−sin⁡2θ=0⇔sin⁡2θ=x21+x2 \begin{aligned} x^2 = \frac{\sin^2\theta}{1-\sin^2\theta} \Leftrightarrow x^2 -x^2\sin^2\theta - \sin^2\theta = 0 \Leftrightarrow \sin^2\theta = \frac{x^2}{1+x^2} \end{aligned}

故（可能需要分类讨论？）

sin⁡θ=x1+x2 \sin\theta = \frac{x}{\sqrt{1+x^2}}

得原积分

−12ln⁡(x+1+x2)1+x2+x21+x2 -\frac{1}{2}\frac{\ln(x+\sqrt{1+x^2})}{1+x^2} + \frac{x}{2\sqrt{1+x^2}}