Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$$
\int \frac{x^2+1}{x\left(x^2-1\right)} \mathrm{d} x=
$$
MathematicsIndefinite IntegrationMHT CETMHT CET 2023 (12 May Shift 1)
Options:
  • A $\log x\left(x^2-1\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • B $\log \left(\frac{x^2-1}{x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • C $\log \left(x^2-1\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • D $\log \left(\frac{x^2+1}{x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
Solution:
1066 Upvotes Verified Answer
The correct answer is: $\log \left(\frac{x^2-1}{x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
$\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{x^2+1}{x\left(x^2-1\right)} \mathrm{d} x \\ & =\int \frac{\frac{x^2+1}{x^2-1}}{x} \mathrm{~d} x \\ & \text { Let } \mathrm{t}=\frac{x^2-1}{x} \Rightarrow \mathrm{dt}=\frac{x^2+1}{x^2} \mathrm{~d} x \\ & \therefore \quad \mathrm{I}=\int \frac{1}{\mathrm{t}} \mathrm{dt}=\log (\mathrm{t})+\mathrm{c}=\log \left(\frac{x^2-1}{x}\right)+\mathrm{c}\end{aligned}$

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.