Question

Evaluate $\int (\ln x{)}^x\left(\ln \right(\ln x)+\frac{1}{\ln x})dx$
(where $C$ is constant of integration)

• A. $(\ln x{)}^x+C$
• B. $-(\ln x{)}^x+C$
• C. $(\ln x{)}^{-x}+C$
• D. $-(\ln x{)}^{-x}+C$

Answer 1

The correct option is A $(\ln x{)}^x+C$

Let $I=\int (\ln x{)}^x\left(\ln \right(\ln x)+\frac{1}{\ln x})dx$
Put $(\ln x{)}^x=t$

$\Rightarrow (\ln x{)}^x\left(\frac{d}{dx}\right(x\ln \left(\ln x\right)\left)\right)dx=dt$

$\Rightarrow (\ln x{)}^x\left(\ln \right(\ln x)+\frac{x}{\ln x}\cdot \frac{1}{x})dx=dt$
$\Rightarrow (\ln x{)}^x\left(\ln \right(\ln x)+\frac{1}{\ln x})dx=dt$

$\therefore I=\int dt=t+C=(\ln x{)}^x+C$