MHT CET · Maths · Indefinite Integration
\( \int \frac{x^2+1}{x\left(x^2-1\right)} \mathrm{d} x= \)
- 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.
Answer & Solution
Correct Answer
(B) \(\log \left(\frac{x^2-1}{x}\right)+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{x^2+1}{x\left(x^2-1\right)} \mathrm{d} x \\ & =\int \frac{\frac{x^2+1}{x^2-1}}{x} \mathrm{~d} x \\ & \text { Let } \mathrm{t}=\frac{x^2-1}{x} \Rightarrow \mathrm{dt}=\frac{x^2+1}{x^2} \mathrm{~d} x \\ & \therefore \quad \mathrm{I}=\int \frac{1}{\mathrm{t}} \mathrm{dt}=\log (\mathrm{t})+\mathrm{c}=\log \left(\frac{x^2-1}{x}\right)+\mathrm{c}\end{aligned}\)
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