Question

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

Answer 1

$I=\int \left(\log \right(\log x)+\frac{1}{{\left(\log x\right)}^2})dx=\int \log \left(\log x\right)dx+\int \frac{1}{{\left(\log x\right)}^2}dxLet{I}_1=\int \log \left(\log x\right)dx=\int \log \left(\log x\right).1dx\text{continue with integration by part}:\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int \left[f^{\prime }\right(x)\int g\left(x\right)dx]dxnow,I_1=\log \left(\log x\right)\int 1dx-\int \left[\frac{d}{dx}\right(\log \left(\log x\right)).\int 1dx]dx=x\log \left(\log x\right)-\int [\frac{1}{\log x}.\frac{1}{x}.x]dx=x\log \left(\log x\right)-\int \frac{1}{\log x}dxLet{I}_2=\int \frac{1}{\log x}dx=\int \frac{1}{\log x}.1dx=\frac{1}{\log x}.\int 1dx-\int [\frac{d}{dx}\left(\frac{1}{\log x}\right).\int 1dx]dx=\frac{x}{\log x}-\int \frac{-1}{{\left(\log x\right)}^2}.\frac{1}{x}.xdx=\frac{x}{\log x}+\int \frac{1}{{\left(\log x\right)}^2}dxputtingthevalueof{I}_{2}in{I}_1{I}_1=x\log \left(\log x\right)-\frac{x}{\log x}-\int \frac{1}{{\left(\log x\right)}^2}dxnowputtingthevalueof{I}_{1}inII=x\log \left(\log x\right)-\frac{x}{\log x}-\int \frac{1}{{\left(\log x\right)}^2}dx+\int \frac{1}{{\left(\log x\right)}^2}dx=x\log \left(\log x\right)-\frac{x}{\log x}+C,whereC=integrationconstant$