Evaluate $\int e^{\tan^{-1} x} \left( \frac{1+x+x^2}{1+x^2} \right) dx$
Step-by-Step Solution
Key Concept: General
$\int e^{\tan^{-1} x} \left( \frac{1+x^2}{1+x^2} + \frac{x}{1+x^2} \right) dx = \int e^{\tan^{-1} x} \left( 1 + \frac{x}{1+x^2} \right) dx = \int (\underbrace{e^{\tan^{-1} x}}_{f(x)} + \underbrace{x \cdot \frac{e^{\tan^{-1} x}}{1+x^2}}_{x f'(x)}) dx = x e^{\tan^{-1} x} + C$
Correct Answer: $x e^{\tan^{-1} x} + C$