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: Use trigonometric inverse identities.
Let \[ t=\tan^{-1}x \] Then, \[ \frac{dt}{dx}=\frac{1}{1+x^2} \] Also, \[ \sec t=\sqrt{1+\tan^2t} \] Since \[ \tan t=x, \] we get \[ \sec t=\sqrt{1+x^2} \] Therefore, \[ \sec^{-1}\sqrt{1+x^2}=t=\tan^{-1}x \] Also, \[ \cos 2t=\frac{1-\tan^2t}{1+\tan^2t} \] So, \[ \cos 2t=\frac{1-x^2}{1+x^2} \] Hence, \[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2t=2\tan^{-1}x \]

Step 2: Convert the integral in terms of \(t\).
The given integral becomes \[ \int e^t\left(t^2+2t\right)dt \]

Step 3: Recognize the derivative form.
We know that \[ \frac{d}{dt}\left(e^t t^2\right) = e^t t^2+e^t\cdot 2t \] \[ = e^t(t^2+2t) \] Therefore, \[ \int e^t(t^2+2t)dt=e^t t^2+C \]

Step 4: Substitute back \(t=\tan^{-1}x\).
Thus, \[ e^t t^2+C = e^{\tan^{-1}x}\left(\tan^{-1}x\right)^2+C \]

Step 5: Final conclusion.
Therefore, \[ \boxed{e^{\tan^{-1}x}\left(\tan^{-1}x\right)^2+C} \]