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
- A \(e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^2+c\)
- B \(e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)+c\)
- C \(e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^3+c\)
- D \(-e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^2+c\)
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 \(\tan ^{-1} x=\theta \Rightarrow x=\tan \theta\)…
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