Question

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

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

Answer 1

The Correct Option is C

Let $I=\int \frac{1+\log x}{(1+x \log x)^{2}} d x$ Let $1+x \log x=t$ $\Rightarrow\left(0+x \times \frac{1}{x}+\log x\right) d x=d t$ $\Rightarrow (1+\log x) d x=d t$ $\therefore I=\int \frac{1}{t^{2}} d t=\frac{t^{-1}}{-1}+C$ $=\frac{-1}{(1+x \log x)}+C$