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CUET · MATHS · PYQ PAPER 2023

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

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

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\frac{d}{dx}\left(x e^{\tan^{-1}x}\right) = e^{\tan^{-1}x} \cdot 1 + x \cdot e^{\tan^{-1}x} \cdot \frac{1}{1+x^2} = e^{\tan^{-1}x} \left(\frac{1+x^2+x}{1+x^2}\right)\) \(\therefore \int\left(\frac{1+x+x^2}{1+x^2}\right) e^{\tan ^{-1} x} d x = x e^{\tan ^{-1} x} + c\)
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