Search any question & find its solution
Question:
Answered & Verified by Expert
$\int \frac{(\log x-1)^2}{\left[1+(\log x)^2\right]^2} d x=$ (where $C$ is constant of integration.)
Options:
Solution:
2582 Upvotes
Verified Answer
The correct answer is:
$\frac{x}{1+(\log x)^2}+C$
$\begin{aligned} & \int \frac{(\log x-1)^2}{\left\{1+(\log x)^2\right\}^2} \mathrm{~d} x=\int \frac{(\log x)^2+1-2 \log x}{\left\{1+(\log x)^2\right\}^2} \mathrm{~d} x \\ & =\int\left\{\frac{1}{1+(\log x)^2}-\frac{2 \log x}{\left\{1+(\log x)^2\right\}^2}\right\} \mathrm{d} x \\ & \int\left\{x\left\{\frac{-2 \log x}{x \cdot\left(1+(\log x)^2\right)^2}\right\}+\frac{1}{1+(\log x)^2}\right\} \mathrm{d} x \\ & =\frac{x}{1+(\log x)^2}+C\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.