Question

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

Answer 1

$\text{Let}I=\int \left[\log \right(\log x)+\frac{1}{(\log x{)}^2}]dx$

$\text{Let}\log x=t\Rightarrow x=e^t\Rightarrow dx=e^{t}dt\therefore I=\int \{\log t+\frac{1}{t^2}\}{e}^{t}dt=\int \{\log t+\frac{1}{t}-\frac{1}{t}+\frac{1}{t^2}\}{e}^{t}dt=\int (\log t+\frac{1}{t})e^t+(-\frac{1}{t}+\frac{1}{t^2})e^t=e^t\cdot \log t-\frac{1}{t}\cdot e^t+C[\because \int \left(f\right(x)+f^{\prime }(x\left)\right)e^{x}dx=f(x)e^x+C]$

$\text{Put}t=\log x=e^{\log x}\log \left(\log x\right)-\frac{1}{\log x}{e}^{\log x}+C\Rightarrow x\cdot \log \left(\log x\right)-\frac{x}{\log x}+C$