Question

Solve $\int \left[\log \left(logx\right)+\frac{1}{\left(\log x\right)}\right]dx$

• A. $I=4x\log \left(\log x\right)+C$
• B. $I=x\log \left(\log x\right)+C$
• C. $I=\log \left(\log x\right)+C$
• D. None of these

Answer 1

Answer: (B)

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

Consider the given integral.

$I=[\int \log \left(\log x\right)+\frac{1}{\log x}]dx$

We know that

$\int uvdx=u\int vdx-\int (\frac{d}{dx}\left(u\right)\int vdx)dx$

Therefore,

$I=[\int \log \left(\log x\right)+\frac{1}{\log x}]dx$

$I=\int \log \left(\log x\right)x-\int \frac{1}{\log x}\times \frac{1}{x}\times xdx+\int \frac{1}{\log x}dx$

$I=x\log \left(\log x\right)-\int \frac{1}{\log x}dx+\int \frac{1}{\log x}dx$

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