Question

$\int \frac{1+\log x}{(1+x\log x{)}^2}dx$ is equal to :

• A. $\frac{1}{1+x\log |x|}+C$
• B. $\frac{1}{1+\log |x|}+C$
• C. $\frac{-1}{1+x\log |x|}+C$
• D. $\log \left|\frac{1}{1+\log |x|}\right|+C$
• E. $\log |1+x\log |x||+C$

Answer 1

The correct option is C $\frac{-1}{1+x\log |x|}+C$

Let $I=\int \frac{1+\log x}{(1+x\log x{)}^2}dx$
Let $1+x\log x=t$
$\Rightarrow (0+x\times \frac{1}{x}+\log x)dx=dt$
$\Rightarrow (0+x\times \frac{1}{x}+\log x)dx=dt$
$\Rightarrow (1+\log x)dx=dt$
Therefore, $I=\int \frac{1}{t^2}dt=\frac{t^{-1}}{-1}+C$
On resubstituting value of $t$, we get

$I=\frac{-1}{(1+x\log x)}+C$