Question

Evaluate the integral \( \int \frac{e^{\tan^{-1} x}}{1 + x^2} \, dx \)

• A. \( e^{\tan^{-1} x} + c \)
• B. \( x e^{\tan^{-1} x} + c \)
• C. \( e^{\tan^{-1} x} \left( 1 + x^2 \right) + c \)
• D. \( x e^{\tan^{-1} x} \left( 1 + x^2 \right) + c \)

Hint:

When evaluating integrals involving \( \tan^{-1} x \), use substitution to simplify the expression.

Answer 1

The Correct Option is B

Step 1: Substitution.
Let \( \theta = \tan^{-1} x \), so that \( \tan \theta = x \). Also,
\[ \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2} \]
Thus, the given integral can be written as: \[ \int e^{\theta} \, d\theta \]

Step 2: Integrating with respect to \( \theta \).
The integral of \( e^{\theta} \) is simply \( e^{\theta} \), so we get:
\[ \int e^{\theta} \, d\theta = e^{\theta} + c \]

Step 3: Replacing \( \theta \) back in terms of \( x \).
Since \( \theta = \tan^{-1} x \), we replace \( \theta \) back in the result:
\[ e^{\theta} = e^{\tan^{-1} x} \]

Step 4: Conclusion.
Therefore, the integral becomes:
\[ \int \frac{e^{\tan^{-1} x}}{(1 + x^2)} \, dx = x e^{\tan^{-1} x} + c \]
So, the correct answer is option (B).