Search any question & find its solution
Question:
Answered & Verified by Expert
$\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x=$
Options:
Solution:
2872 Upvotes
Verified Answer
The correct answer is:
$e^{\tan ^{-1} x}\left(\tan ^{-1} x\right)^2+C$
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
$I=\int e^t\left[t^2+2 t\right] d t \Rightarrow I=\int\left(t^2 e^t+2 e^t t\right) d t$
By integration by parts, we get
$\begin{aligned} & I=t^2 e^t-\int 2 t e^t d t-\int 2 t e^t d t+C \\ & I=t^2 e^t+C=e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^2+C\end{aligned}$
$I=\int \frac{e^{\tan ^{-1}(x)}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{\left.1+x^2\right)^2}+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x\right.$
On putting $\tan ^{-1} x=t$ and $\frac{d x}{1+x^2}=d t$, we get
$I=\int e^t\left[t^2+2 t\right] d t \Rightarrow I=\int\left(t^2 e^t+2 e^t t\right) d t$
By integration by parts, we get
$\begin{aligned} & I=t^2 e^t-\int 2 t e^t d t-\int 2 t e^t d t+C \\ & I=t^2 e^t+C=e^{\tan ^{-1}(x)}\left(\tan ^{-1} x\right)^2+C\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.