Question

$\int \frac{\log x}{(1+x{)}^3}dx$ is equal to

• A. $\frac{\log x}{(1+x{)}^2}+\frac{1}{2}\log \frac{x}{x+1}+\frac{1}{2}\frac{1}{x+1}+c$
• B. $2\frac{-\log x}{(1+x{)}^2}+2\log \frac{x+1}{x}+\frac{2}{x+1}+c$
• C. $\frac{-2\log x}{(1+x{)}^2}+2\log \frac{x}{x+1}+\frac{2}{x+1}+c$
• D. $\frac{-\log x}{(1+x{)}^2}+\frac{1}{2}\log \frac{x+1}{x}-\frac{1}{2}\frac{1}{x+1}+c$

Answer 1

Answer: (C)

$\frac{-2\log x}{(1+x{)}^2}+2\log \frac{x}{x+1}+\frac{2}{x+1}+c$

$I=\int \frac{\log x}{(1+x{)}^3}dx$
$=\log x\int \frac{1}{(1+x{)}^3}dx-\int \frac{-2}{x(1+x{)}^2}dx$
$=\frac{-2\log x}{(1+x{)}^2}+2\int \frac{1}{x(1+x{)}^2}dx$
Now, $\frac{1}{x(1+x{)}^2}=\frac{A}{x}+\frac{B}{(1+x)}+\frac{C}{(1+x{)}^2}$
$\Rightarrow 1=A(1+x{)}^2+Bx(1+x)+Cx$
When $x=-1,C=-1$
When $x=0,A=1$
When $x=1,B=-1$
So $\frac{1}{x(1+x{)}^2}=\frac{1}{x}-\frac{1}{(1+x)}-\frac{1}{(1+x{)}^2}$
$\int \frac{1}{x(1+x{)}^2}dx=\int \frac{1}{x}dx-\int \frac{dx}{(1+x)}-\int \frac{dx}{(1+x{)}^2}$
$=\log x-\log (x+1)+\frac{1}{(x+1)}$
So, $I=\frac{-2\log x}{(1+x{)}^2}+2\log x-2\log (x+1)+\frac{2}{(x+1)}+C$
$I=\frac{-2\log x}{(1+x{)}^2}+2\log \frac{x}{x+1}+\frac{2}{(x+1)}+C$