Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \left[\log \left(\log x\right)+\frac{1}{{\left(\log x\right)}^2}\right]dx \\ \mathrm{Putting}\log x=t \\ \Rightarrow x=e^t \\ \Rightarrow dx=e^{t}dt \\ \therefore I=\int \left[\log \left(t\right)+\frac{1}{t^2}\right]e^{t}dt \\ =\int \left(\log t+\frac{1}{t}-\frac{1}{t}+\frac{1}{t^2}\right)e^t \\ =\int \left(\log t+\frac{1}{t}\right)e^t+\int \left(-\frac{1}{t}+\frac{1}{t^2}\right)e^{t}dt \\ =e^t\log t+e^t\left(-\frac{1}{t}\right)+C\left[\because \int e^x\left\{f\left(x\right)+f\text{'}\left(x\right)\right\}dx=e^{x}f\left(x\right)+C\right] \\ =e^t\left(\log t-\frac{1}{t}\right)+C \\ =e^{\log x}\left[\log \left(\log x\right)-\frac{1}{\log x}\right]+C \\ =x\left[\log \left(\log x\right)-\frac{1}{\log x}\right]+C \end{gathered}$$