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COMEDK · Maths · 28. Indefinite Integration

\(\int \dfrac{e^{\log\left(1+\frac{1}{x^2}\right)}}{x^2 + \dfrac{1}{x^2}}\ dx =\)

  1. A \(\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x^2-1}{\sqrt{2}\,x}\right) + C\)
  2. B \(\dfrac{1}{\sqrt{2}}\tan^{-1}\left(x - \dfrac{1}{x}\right) + C\)
  3. C \(\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x^2+1}{x\sqrt{2}}\right) + C\)
  4. D \(-\dfrac{1}{\sqrt{2}}\tan^{-1}\left(x - \dfrac{1}{x}\right) + C\)
Verified Solution

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…