ExamBro
ExamBro
AP EAMCET · Maths · Indefinite Integration

The value of \(\int \frac{e^{\tan ^{-1}(x)}}{1+x^2}\) \(\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x\), for \(x>0\) is

  1. A \(e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^2+c\)
  2. B \(e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)+c\)
  3. C \(e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^3+c\)
  4. D \(-e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^2+c\)
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 \(I=\int \frac{e^{\tan ^{-1} x}}{1+x^2}\) \[ \left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x(x>0) \] Let us take \(\tan ^{-1} x=\theta \Rightarrow x=\tan \theta\)…