Question

$\int [\log \left(\log x\right)+\frac{1}{{\left(\log x\right)}^2}]dx.$

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

Answer 1

The correct option is B $x\log \left(\log x\right)-\frac{x}{\log x}$

Put $\log x=t\therefore x=e^t$ and $dx=e^{t}dt.$
$I=\int e^t(\log t+\frac{1}{t^2})dt=\int e^t(\log t+\frac{1}{t}-\frac{1}{t}+\frac{1}{t^2})dt$
$=e^t(\log t+\frac{1}{t})dt+\int e^t(-\frac{1}{t}+\frac{1}{t^2})dt$
$=e^t\log t+e^t(-\frac{1}{t})=x\log \left(\log x\right)-\frac{x}{\log x}$