$\int \frac{x^2+1}{x\left(x^2-1\right)} \mathrm{d} x=$

$\int \frac{x^2+1}{x\left(x^2-1\right)} \mathrm{d} x=$
  1. $\log x\left(x^2-1\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  2. $\log \left(\frac{x^2-1}{x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  3. $\log \left(x^2-1\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  4. $\log \left(\frac{x^2+1}{x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.

Solution

$\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{x^2+1}{x\left(x^2-1\right)} \mathrm{d} x \\ & =\int \frac{\frac{x^2+1}{x^2-1}}{x} \mathrm{~d} x \\ & \text { Let } \mathrm{t}=\frac{x^2-1}{x} \Rightarrow \mathrm{dt}=\frac{x^2+1}{x^2} \mathrm{~d} x \\ & \therefore \quad \mathrm{I}=\int \frac{1}{\mathrm{t}} \mathrm{dt}=\log (\mathrm{t})+\mathrm{c}=\log \left(\frac{x^2-1}{x}\right)+\mathrm{c}\end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 1)

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