Question

Solve :

$\int \frac{1}{x+x\log x}dx.$

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

Answer 1

The correct option is C $\log (1+\log x).$

Let $I=\int \frac{1}{x+x\log x}dx=\int \frac{1}{x(1+\log x)}dx$

Put $1+\log x=t\Rightarrow \frac{1}{x}dx=dt$

Therefore

$I=\int \frac{dt}{t}=\log t=\log (1+\log x)$