Question

$\int \left(\frac{1}{x}\right)\log xdx=$

• A. ${\log }_e\left(1-{\log }_{e}x\right)+C$
• B. ${\log }_e\left({\log }_{e}ex-1\right)+C$
• C. $\log \left(\log x-1\right)+C$
• D. ${\log }_e\left({\log }_{e}x+x\right)+C$
• E. $\log \left(\log x\right)+C$

Answer 1

The correct option is E

$\log \left(\log x\right)+C$

Explanation for correct answer:

Solving this by using the substitution method

$$\begin{gathered} t=\log x\Rightarrow dt=\frac{1}{x}dx...\left[\because \frac{d}{dx}logx=\frac{1}{x}\right] \\ \int \left(\frac{1}{x}\right)\log xdx=\int t.dt \\ =\frac{t^2}{2} \\ =\frac{1}{2}{\left(\log x\right)}^2+C \\ =\log x+C...\left[\because {(loga)}^n=nloga\right] \end{gathered}$$

$$\begin{gathered} \int \left(\frac{1}{x\log x}\right)dx=\int \frac{1}{t}.dt \\ =\log t+c \\ =\log \left(\log x\right)+C \end{gathered}$$

Hence, option (E) is the correct answer.