Question

Find the value of \[ \int e^x \left( \tan^{-1} x + \frac{1}{1 + x^2} \right) dx. \] 

Hint:

For integrals involving both exponential and trigonometric functions, consider breaking them into simpler parts or using integration by parts.

Answer 1

Step 1: Break the integral into two parts: \[ \int e^x \tan^{-1} x \, dx + \int e^x \frac{1}{1 + x^2} \, dx. \] Step 2: The second integral is straightforward: \[ \int e^x \frac{1}{1 + x^2} \, dx = e^x \cdot \tan^{-1} x + C. \] Step 3: For the first integral, use integration by parts or refer to standard integral tables for \( \int e^x \tan^{-1} x \, dx \). The general form will involve \( e^x \) and the arctangent function. Thus, the answer is a combination of these integrals.