Search any question & find its solution
Question:
Answered & Verified by Expert
$\int \frac{\left(\mathrm{x}^{2}-1\right)}{\left(\mathrm{x}^{2}+1\right) \sqrt{\mathrm{x}^{4}+1}} \mathrm{dx}$ is equal to
Options:
Solution:
1349 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{x^{2}+1}{\sqrt{2} x}\right)+c$
$\mathrm{I}=\int \frac{\mathrm{x}^{2}\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{dx}}{\mathrm{x}^{2}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\left(\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right)^{1 / 2}}$
Let $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{p} \Rightarrow\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}=\mathrm{dp}$
$\mathrm{I}=\int \frac{\mathrm{dp}}{\mathrm{p} \sqrt{\mathrm{p}^{2}-2}}=\frac{1}{\sqrt{2}} \mathrm{sec}^{-1} \frac{\mathrm{p}}{\sqrt{2}}$
$=\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{\mathrm{x}^{2}+1}{\sqrt{2} \mathrm{x}}\right)+\mathrm{c}$
Let $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{p} \Rightarrow\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}=\mathrm{dp}$
$\mathrm{I}=\int \frac{\mathrm{dp}}{\mathrm{p} \sqrt{\mathrm{p}^{2}-2}}=\frac{1}{\sqrt{2}} \mathrm{sec}^{-1} \frac{\mathrm{p}}{\sqrt{2}}$
$=\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{\mathrm{x}^{2}+1}{\sqrt{2} \mathrm{x}}\right)+\mathrm{c}$
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.