Integrals - Integral of the type ∫ e^x(f(x) + f'(x)) dx

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

Let $f(x) = \tan^{-1} x$. Then, we know that $f'(x) = \frac{1}{1+x^2}$. The integral is in the form $\int e^x [f(x) + f'(x)] dx$. Substituting the values, we get: $I = e^x \tan^{-1} x + C$

Direct application of the formula where $f(x)$ is the inverse trigonometric function and $f'(x)$ is its standard derivative.

Evaluate $\int \frac{x e^x}{(1+x)^2} dx$.

We rewrite the numerator $x$ as $(x + 1) - 1$: $I = \int e^x \left[ \frac{x+1-1}{(1+x)^2} \right] dx$ $I = \int e^x \left[ \frac{x+1}{(1+x)^2} - \frac{1}{(1+x)^2} \right] dx$ $I = \int e^x \left[ \frac{1}{1+x} + \left( -\frac{1}{(1+x)^2} \right) \right] dx$ Let $f(x) = \frac{1}{1+x}$. Then $f'(x) = -\frac{1}{(1+x)^2}$. Using the property, we get: $I = \frac{e^x}{1+x} + C$

This example requires algebraic manipulation to split the fraction into a function and its derivative.

Evaluate $\int e^x (\log x + \frac{1}{x}) dx$.

Let $f(x) = \log x$. We know that the derivative $f'(x) = \frac{1}{x}$. The integral matches the form $\int e^x [f(x) + f'(x)] dx$. Therefore: $I = e^x \log x + C$

A straightforward case involving logarithmic and power functions.