Question

Evaluate $\int \log \left(\log x\right)dx+\int (\log x{)}^{-2}dx$

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

Answer 1

Answer: (C)

$x\log \left(\log x\right)-\frac{x}{\log x}+c$

Consider, $I=\int \log \left(\log x\right)dx+\int (\log x{)}^{-2}dx$

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

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

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

Now, $\int \frac{1}{\left(\log x\right)}dx=\frac{1}{\log x}\int 1\cdot dx-\int \frac{-1}{(\log x{)}^2}\cdot \frac{1}{x}\cdot xdx$

$=\frac{x}{\log x}+\int \frac{1}{(\log x{)}^2}dx.$

Substituting in $\left(i\right)$, we get

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

$=x\log \left(\log x\right)-\frac{x}{\log x}+c.$

$\therefore \int \log \left(\log x\right)dx+\int (\log x{)}^{-2}dx=x\log (\log x)-\frac{x}{\log x}+c.$