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:

Look for substitution when \( \frac{1}{1+x^2} \) appears: it's typically \( \frac{d}{dx}(\tan^{-1}x) \).

Answer 1

The Correct Option is A

Let: \[ I = \int \frac{e^{\tan^{-1}x}}{1+x^2} \left[\left(\sec^{-1}(\sqrt{1+x^2})\right)^2 + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)\right] dx \] Now observe: - \( \frac{1}{1+x^2} \) is the derivative of \( \tan^{-1}x \)
- Let \( t = \tan^{-1}x \Rightarrow x = \tan t \Rightarrow dx = \sec^2 t dt \)
- Also note:
- \( \sec^{-1}(\sqrt{1+x^2}) = \sec^{-1}(\sqrt{1+\tan^2 t}) = \sec^{-1}(\sec t) = t \)
- So \( \left( \sec^{-1}(\sqrt{1+x^2}) \right)^2 = t^2 \)
- \( \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) = \cos^{-1}(\cos 2t) = 2t \)
So, the expression becomes: \[ I = \int e^t (t^2 + 2t) dt = \int e^t t^2 dt + \int 2t e^t dt \Rightarrow \text{Use integration by parts or shortcut:} \] We know: \[ \int e^t t^2 dt = e^t (t^2 - 2t + 2) + C,\quad \int 2t e^t dt = 2e^t(t - 1) \] Add: \[ e^t (t^2 - 2t + 2) + 2e^t(t - 1) = e^t (t^2 - 2t + 2 + 2t - 2) = e^t t^2 \Rightarrow I = e^{\tan^{-1}x} \cdot (\tan^{-1}x)^2 + C \]