Question

Integrate the function $\frac{1}{x+x\log x}$

• A. $\log (1+\log x)+c$
• B. $\log (1+2\log x)+c$
• C. $\log (2+\log x)+c$
• D. None of these

Answer 1

The correct option is A $\log (1+\log x)+c$

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

$={\int }^{}\frac{1}{x\left(1+\log x\right)}dx$

putting $1+logx=t$

$\Rightarrow \frac{1}{x}dx=dt$

$={\int }^{}\frac{1}{t}dt$

$=\log \left|t\right|+c$

$=\log \left(1+\log x\right)+c$