Search any question & find its solution
Question:
Answered & Verified by Expert
The value of $\int \frac{x^{2}+1}{x^{2}-1} d x$ is
Options:
Solution:
2227 Upvotes
Verified Answer
The correct answer is:
$x+\log \left(\frac{x-1}{x+1}\right)+C$
Let $I=\int \frac{x^{2}+1}{x^{2}-1} d x=\int \frac{x^{2}-1+2}{x^{2}-1} d x$
$$
=\int\left(1+\frac{2}{x^{2}-1}\right) d x=x+2 \times \frac{1}{2} \log \left|\frac{x-1}{x+1}\right|+C
$$
$$
=x+\log \left|\frac{x-1}{x+1}\right|+C
$$
$$
=\int\left(1+\frac{2}{x^{2}-1}\right) d x=x+2 \times \frac{1}{2} \log \left|\frac{x-1}{x+1}\right|+C
$$
$$
=x+\log \left|\frac{x-1}{x+1}\right|+C
$$
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.