Question

Evaluate \( \int e^{\tan^{-1} x} \left( 1 + \frac{x}{1 + x^2} \right) dx \)

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

Hint:

When dealing with integrals involving inverse trigonometric functions, simplify the integrand first, then apply substitution and standard integration techniques.

Answer 1

The Correct Option is B

Step 1: Simplify the expression.
We are given the integral \( \int e^{\tan^{-1} x} \left( 1 + \frac{x}{1 + x^2} \right) dx \). First, simplify the integrand: \[ \left( 1 + \frac{x}{1 + x^2} \right) = \frac{1 + x^2 + x}{1 + x^2} = \frac{(1 + x^2) + x}{1 + x^2} \] This gives: \[ \int e^{\tan^{-1} x} \cdot \frac{(1 + x^2) + x}{1 + x^2} \, dx \]
Step 2: Substitute for \( \tan^{-1} x \).
Now, substitute \( u = \tan^{-1} x \), which simplifies the integral to: \[ \int x e^{u} + C \] Hence, we get: \[ x e^{\tan^{-1} x} + C \]
Step 3: Conclusion.
Thus, the correct answer is \( x e^{\tan^{-1} x} + C \).