Q. ${\int_{}^{}\left\{ \frac{\log{x - 1}}{1 + ({\log{x)\ }}^{2}} \right\}}^{2}dx$ is equal to
A) $\frac{x}{{(\log{x)}}^{2} + 1} + c$
B) $\frac{xe^{x}}{1 + x^{2}} + c$
C) $\frac{x}{x^{2} + 1} + c$
D) $\frac{\log x}{\left( \log x \right)^{2} + 1} + c$
Explanation
(a)
${\int_{}^{}\left\{ \frac{(\log{x - 1})}{1 + \left( \log x \right)^{2}} \right\}}^{2}dx = \int_{}^{}{\frac{({\log{x)}}^{2} + 1 - 2\log x\ }{\left\lbrack ({\log{x)}}^{2} + 1 \right\rbrack^{2}}\ dx}$
$= \int_{}^{}\frac{({\log{x)}}^{2} + 1 - 2x\left( \log{x.\frac{1}{x}} \right)}{\left\lbrack ({\log{x)}}^{2} + 1 \right\rbrack^{2}}\ dx$
$= \int_{}^{}{\frac{d}{dx}(\frac{x}{({\log{x)}}^{2} + 1}}\ dx = \frac{x}{{(\log{x)}}^{2} + 1} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.