Question

Evaluate: $$ \int \frac{e^{\tan^{-1} x}}{1 + x^2} \left[ \left(\sec^{-1}\sqrt{1 + x^2}\right)^2 + \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \right] dx, \quad x > 0 $$ 

• A. $(\tan^{-1}x)^{2}e^{\tan^{-1}x}+c$, where c is a constant of integration.
• B. $(\tan^{-1}x)e^{\tan^{-1}x}+c$, where c is a constant of integration.
• C. $(\tan^{-1}x)e^{2\tan^{-1}x}+c$, where c is a constant of integration.
• D. $(\tan^{-1}x)^{2}e^{2\tan^{-1}x}+c$, where c is a constant of integration.

Hint:

Logic Tip: The terms $\sqrt{1+x^2}$ and $\frac{1-x^2}{1+x^2}$ are classic hallmarks of trigonometric substitution involving $x = \tan \theta$. Recognizing these forms immediately simplifies complex inverse trigonometric functions into basic algebraic terms.

Answer 1

The Correct Option is A