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AP EAMCET · Maths · Indefinite Integration

\(\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x=\)

  1. A \(e^{\tan ^{-1} x}\left(\tan ^{-1} x\right)^2+C\)
  2. B \(e^{\tan ^{-1} x}\left(\sec ^{-1} x\right)^2+C\)
  3. C \(e^{\tan ^{-1} x}\left(\sec ^{-1} \cdot\left(\sqrt{1+x^2}\right)\right)+C\)
  4. D \(e^{\tan ^{-1} x}\left(\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)+C\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(e^{\tan ^{-1} x}\left(\tan ^{-1} x\right)^2+C\)

Step-by-step Solution

Detailed explanation

Let \(I=\int \frac{e^{\tan ^{-1}(x)}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{\left.1+x^2\right)^2}+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x\right.\) On putting \(\tan ^{-1} x=t\) and \(\frac{d x}{1+x^2}=d t\), we get…