Question

Solve:
$\int \frac{x}{{(x+1)}^2}dx$

Answer 1

$\int \frac{x}{{(x+1)}^2}dx$

$=\int \frac{x+1-1}{{(x+1)}^2}dx$

$=\int \frac{(x+1)dx}{{(x+1)}^2}-\int \frac{1}{{(x+1)}^2}dx$

For $\int \frac{(x+1)dx}{{(x+1)}^2},$ put ${(x+1)}^2=t$

$\Rightarrow (x+1)dx=\frac{dt}{2}$

$\Rightarrow \int \frac{1}{t}\frac{dt}{2}=\frac{1}{2}\int \frac{dt}{2}.\frac{1}{2}\log t$

for $\int \frac{1}{{(x+1)}^2}dx,$ $x+1=d\Rightarrow dx=db$

$\Rightarrow \int \frac{1}{b^2}db=\frac{b^{-1}}{-1}=\frac{-1}{b}=\frac{-1}{(x+1)}$

$\therefore \int \frac{x}{{(x+1)}^2}dx=\frac{1}{2}\log (x+1)-\frac{1}{x+1}+C.$

Hence, solved.