COMEDK · Maths · 28. Indefinite Integration
\(\int \dfrac{e^{\tan ^{-1} x}}{\left(1+x^2\right)}\left(1+x+x^2\right) d x=\)
- A \(\dfrac{x e^{\tan ^{-1} x}}{\left(1+x^2\right)}+c\)
- B \(e^{\tan ^{-1} x}+c\)
- C \(x e^{\tan ^{-1} x}+c\)
- D \(\dfrac{e^{\tan ^{-1} x}}{\left(1+x^2\right)}+c\)
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…
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