xtan^(-1)x 的積分
Integral of xtan^(-1)x 求解 $\int xtan^{-1}xdx$ 解: $$I=\int xtan^{-1}xdx$$ $$\;\;\;\;\;=\int tan^{-1}xd(\frac{x^2}{2})$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{x^2}{2}tan^{-1}x-\int \frac{x^2}{2}d(tan^{-1}x)$$ 已知 $$(tanx)^{'} = (\frac{sinx}{cosx})^{'}=\frac{cos^2x+sin^2x}{cos^2x}=sec^2 x$$ 即 $$\frac{d(tanx)}{dx}=sec^2x$$