眞趣盎然集

Integral of x tan^(-1) x 求解 $\int x\tan^{-1}x\;dx$ 解： $$I=\int x\tan^{-1}x\;dx\;$$ $$\;\;\;\;=\int \tan^{-1}x\;d(\frac{x^2}{2})$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{x^2}{2}\tan^{-1}x-\int \frac{x^2}{2}d(\tan^{-1}x)$$ 已知  $$(\tan x)^{'} = (\frac{\sin x}{\cos x})^{'}=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}=\sec^2 x$$ 即 $$\frac{d(\tan x)}{dx}=\sec^2 x$$ 所以， $$d(\tan x)=\sec^2 x\;dx$$ 又 $$\tan \left (\tan^{-1}x \right )=x$$ $$d \left [\tan \left (\tan^{-1}x \right ) \right ]=dx$$ $$\sec^2 \left (\tan^{-1}x \right )d \left (\tan^{-1}x \right )=dx$$ $$\left [1+\tan^2 \left (\tan^{-1}x \right )\right]d(\tan^{-1}x)=dx$$ $$(1+x^2)d(\tan^{-1}x)=dx$$ 可得 $$d \left (tan^{-1}x \right )=\frac{dx}{1+x^2}$$ 因此， $$I=\frac{x^2}{2}\tan^{-1}x-\int \frac{x^2}{2}\frac{dx}{1+x^2}$$ $$\;\;\;\;\;=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2} \int \frac{x^2}{1+x^2}dx$$ $$\;\;\;\;\;\;\;\;\;\;\;=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2} \int \frac{1+x^2-1}{1+x^2}dx$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\...