Question

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

Answer 1

Let $I=\int \left[\log \right(\log x)+\frac{1}{(\log x{)}^2}]dx$
$\Rightarrow I=\int \log \left(\log x\right)dx+\int \frac{1}{(\log x{)}^2}dx$
Let$I_1=\int \left[\log \right(\log x\left)\right]dx$
Using integration by parts
$\Rightarrow I_1=\log \left(\log x\right)\int 1dx-\int \left[\frac{d}{dx}\right(\log \left(\log x\right))\int 1dx]dx$
$\Rightarrow I_1=x!\log \left(\log x\right)-\int [\frac{1}{x\log x}.x]dx$
Let $I_2=\int \frac{1}{\log x}dx$
Using itegration by parts
$\Rightarrow I_2=\frac{1}{\log x}\int 1dx-\int [\frac{d\left(\frac{1}{\log x}\right)}{dx}\int 1dx]dx$
$\Rightarrow I_2=\frac{x}{\log x}-\int [-\frac{1}{x(\log x{)}^2}]dx$
$\Rightarrow I_1=x\log \left(\log x\right)-\frac{x}{\log x}-\int \frac{1}{(logx{)}^2}dx$
$I=x\log \left(\log x\right)-\frac{x}{\log x}+C$
Where $C$ is constant of integration.
Hence, $\int \left[\log \right(\log x)+\frac{1}{(\log x{)}^2}]dx=x\log \left(\log x\right)-\frac{x}{\log x}+C$