填4
\(\displaystyle \int \frac{ln(x+1)}{x^2}dx=\) 。
[解答]
\(u= ln(x+1),dv=x^{-2} dx\)
因此\(uv-\int v du =\displaystyle \frac{-ln(x+1)}{x}+\int \frac{1}{x(x+1)} dx =\displaystyle \frac{-ln(x+1)}{x}+\int \frac{1}{x} dx -\int \frac{1}{x+1} dx \)
所求為\(\displaystyle \frac{-ln(x+1)}{x}+ln|x|-ln|x+1|+C\)
