Question

Evaluate \[ \int \frac{e^{\tan^{-1}x}}{1+x^2} \left[ \left(\sec^{-1}\sqrt{1+x^2}\right)^2+ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \right]dx \]

• 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. \(e^{\tan^{-1}x}\left(\sec^{-1}\sqrt{1+x^2}\right)+C\)
• D. \(e^{\tan^{-1}x}\left[\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]+C\)

Hint:

Use the identities \[ \sec^{-1}\sqrt{1+x^2}=\tan^{-1}x \] and \[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2\tan^{-1}x \] to simplify inverse trigonometric integrals.

Answer 1

The Correct Option is A

Step 1: Simplify $\sec^{-1}\sqrt{1+x^2}$.
Let $x = \tan\theta$ where $\theta = \tan^{-1}x$. Then $\sqrt{1+x^2} = \sec\theta$, so $\sec^{-1}(\sqrt{1+x^2}) = \sec^{-1}(\sec\theta) = \theta = \tan^{-1}x$.
Step 2: Simplify $\cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right)$.
Using the identity: if $x = \tan\phi$, then $\frac{1-\tan^2\phi}{1+\tan^2\phi} = \cos(2\phi)$. So $\cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}x$.
Step 3: Substitute into the integral.
The bracket becomes $(\tan^{-1}x)^2 + 2\tan^{-1}x$. Let $t = \tan^{-1}x$, so $dt = \frac{dx}{1+x^2}$. The integral becomes $\int e^t(t^2 + 2t)\,dt$.
Step 4: Recognize the pattern $\int e^t[f(t)+f'(t)]\,dt = e^t f(t) + C$.
Let $f(t) = t^2$, then $f'(t) = 2t$. So $t^2 + 2t = f(t) + f'(t)$. By the standard result: $\int e^t(t^2+2t)\,dt = e^t \cdot t^2 + C$.
Step 5: Substitute back $t = \tan^{-1}x$.
$= e^{\tan^{-1}x}(\tan^{-1}x)^2 + C$.
Step 6: State the final answer.
\[\boxed{e^{\tan^{-1}x}(\tan^{-1}x)^2 + C}\]