AP EAMCET · Maths · Indefinite Integration
The value of \(\int e^{\tan ^{-1} x} \cdot \frac{\left(1+x+x^2\right)}{1+x^2} d x\) is
- A \(\tan ^{-1} x+c\)
- B \(e^{\tan ^{-1} x}+c\)
- C \(e^{\tan ^{-1} x}-x+c\)
- D \(x e^{\tan ^{-1} x}+c\)
Answer & Solution
Correct Answer
(D) \(x e^{\tan ^{-1} x}+c\)
Step-by-step Solution
Detailed explanation
Let \(\begin{aligned} & I=\int\left(e^{\tan ^{-1} x} \frac{\left(1+x^2\right)}{\left(1+x^2\right)} d x+e^{\tan ^{-1} x} \cdot \frac{x}{\left(1+x^2\right)}\right) \\ & I=\int\left(e^{\tan ^{-1} x}+e^{\tan ^{-1} x} \frac{x}{1+x^2}\right) d x\end{aligned}\) As we know,…
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