ExamBro
ExamBro
COMEDK · Maths · 28. Indefinite Integration

\(\int \dfrac{e^{\tan ^{-1} x}}{\left(1+x^2\right)}\left(1+x+x^2\right) d x=\)

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

Answer & Solution

Correct Answer

(C) \(x e^{\tan ^{-1} x}+c\)

Step-by-step Solution

Detailed explanation

Let \(I = \int \dfrac{e^{\tan^{-1} x}}{1+x^2} (1+x+x^2) dx\). Substitute \(t = \tan^{-1} x\), so \(dt = \dfrac{1}{1+x^2} dx\). Then \(x = \tan t\). The integral becomes \(I = \int e^t (1 + \tan t + \tan^2 t) dt\). Using the identity \(1 + \tan^2 t = \sec^2 t\), we have…