COMEDK · Maths · 28. Indefinite Integration
\(\int \dfrac{e^{\log\left(1+\frac{1}{x^2}\right)}}{x^2 + \dfrac{1}{x^2}}\ dx =\)
- A \(\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x^2-1}{\sqrt{2}\,x}\right) + C\)
- B \(\dfrac{1}{\sqrt{2}}\tan^{-1}\left(x - \dfrac{1}{x}\right) + C\)
- C \(\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x^2+1}{x\sqrt{2}}\right) + C\)
- D \(-\dfrac{1}{\sqrt{2}}\tan^{-1}\left(x - \dfrac{1}{x}\right) + C\)
Answer & Solution
Correct Answer
(A) \(\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x^2-1}{\sqrt{2}\,x}\right) + C\)
Step-by-step Solution
Detailed explanation
Given integral is \(I = \int \dfrac{e^{\log\left(1+\dfrac{1}{x^2}\right)}}{x^2 + \dfrac{1}{x^2}}\ dx\) Using the property \(e^{\log f(x)} = f(x)\), the integral simplifies to: \(I = \int \dfrac{1+\dfrac{1}{x^2}}{x^2 + \dfrac{1}{x^2}}\ dx\) The denominator can be rewritten by…
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