AP EAMCET · Maths · Indefinite Integration
\(\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x=\)
- A \(e^{\tan ^{-1} x}\left(\tan ^{-1} x\right)^2+C\)
- B \(e^{\tan ^{-1} x}\left(\sec ^{-1} x\right)^2+C\)
- C \(e^{\tan ^{-1} x}\left(\sec ^{-1} \cdot\left(\sqrt{1+x^2}\right)\right)+C\)
- D \(e^{\tan ^{-1} x}\left(\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)+C\)
Answer & Solution
Correct Answer
(A) \(e^{\tan ^{-1} x}\left(\tan ^{-1} x\right)^2+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{e^{\tan ^{-1}(x)}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{\left.1+x^2\right)^2}+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x\right.\) On putting \(\tan ^{-1} x=t\) and \(\frac{d x}{1+x^2}=d t\), we get…
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