Question

Evaluate: $\int lnx\cdot \frac{1}{(x+1{)}^2}dx$ equals

Answer 1

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

Using Integration by parts,we get

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

$=lnx\left(\frac{-1}{x+1}\right)-\int \frac{1}{x}\left(\frac{-1}{x+1}\right)dx$

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

$=\frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$ (Partial fraction)

$1=A(x+1)+Bx$

put $x=0$ we get $A=1$

put $x=-1$ ,we get $B=-1$

$=\frac{-lnx}{x+1}+\int \frac{dx}{x}-\int \frac{dx}{x+1}$

$=\frac{-lnx}{x+1}+lnx-ln(x+1)$

$=\frac{-lnx}{x+1}+ln\left(\frac{x}{x+1}\right)+C$

$\int \frac{lnx}{(x+1{)}^2}dx=\frac{-lnx}{x+1}+ln\left(\frac{x}{x+1}\right)+C$