Đáp án: $\int^2_1f(x)dx=\ln \left(3\right)-3\ln \left(2\right)$
Giải thích các bước giải:
Ta có :
$f(x)=\int f'(x)dx$
$\to f(x)=\int \dfrac{x-1}{x(x+1)^2}dx$
$\to f(x)=\int \dfrac{-(x+1)^2+x(x+1)+2x}{x(x+1)^2}dx$
$\to f(x)=\int -\dfrac1x+\dfrac1{x+1}+\dfrac{2}{(x+1)^2}dx$
$\to f(x)=-\ln \left|x\right|+\ln \left|x+1\right|-\dfrac{2}{x+1}+C$
Mà $f(1)=-1$
$\to f(1)=-\ln \left|1\right|+\ln \left|1+1\right|-\dfrac{2}{1+1}+C=-1$
$\to C=-\ln 2$
$\to f(x)=-\ln \left|x\right|+\ln \left|x+1\right|-\dfrac{2}{x+1}-\ln2$
$\to \int f(x)dx=\int -\ln \left|x\right|+\ln \left|x+1\right|-\dfrac{2}{x+1}-\ln2dx$
$\to \int f(x)dx=-\int \ln \left|x\right|dx+\int \ln \left|x+1\right|dx-\int \dfrac{2}{x+1}dx-\int \ln \left(2\right)dx$
$\to \int f(x)dx=-\left(x\ln \left|x\right|-x\right)+x\ln \left|x+1\right|-x-1+\ln \left|x+1\right|-2\ln \left|x+1\right|-\ln \left(2\right)x+C$ (dùng tích phân từng phần)
$\to \int f(x)dx=-\ln \left(2\right)x-x\ln \left|x\right|+x\ln \left|x+1\right|-\ln \left|x+1\right|-1+C$
$\to \int^2_1f(x)dx=\ln \left(3\right)-3\ln \left(2\right)$
