Question

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

• A. $\ln \left|\frac{1+\ln x}{\ln x}\right|+C$
• B. $x\cdot \ln \left|\frac{\ln x}{1+\ln x}\right|+C$
• C. $\ln \left|\frac{\ln x}{1+\ln x}\right|+C$
• D. $x\cdot \ln \left|\frac{1+\ln x}{\ln x}\right|+C$

Answer 1

The correct option is C $\ln \left|\frac{\ln x}{1+\ln x}\right|+C$

$I=\int \frac{1}{x\cdot \ln x(1+\ln x)}dx$
Putting $\ln x=t$
$\Rightarrow \frac{1}{x}dx=dt\therefore \int \frac{1}{t(1+t)}dt=\int \frac{(1+t)-t}{t(1+t)}dt=\int (\frac{1}{t}-\frac{1}{1+t})dt=\ln |t|-\ln |1+t|+C=\ln \left|\frac{t}{1+t}\right|+C=\ln \left|\frac{\ln x}{1+\ln x}\right|+C$