Example 19: $\int x\,e^x\,dx$

Basics you need

Solution

1. Choose the first and second functions

   $$f(x) = x \ \ (\text{first}), \qquad g(x) = e^x \ \ (\text{second}).$$

   Then $f'(x) = 1$ and $\int g(x)\,dx = e^x$⁠.

   Differentiating $x$ gives 1, so the $x$ disappears; integrating $e^x$ gives $e^x$ again, no harder. The other way round, $\int x\,dx = \frac{x^2}{2}$ would raise the power and give $\frac{x^2}{2}e^x - \int\frac{x^2}{2}e^x\,dx$⁠, a harder integral (Example 17 shows the same effect for $\int x\cos x\,dx$⁠).
   Remark (iii) after Example 17: a power of $x$ is usually the first function. Lesson: integration by parts

2. Apply integration by parts

   $$\begin{aligned} \int x\,e^x\,dx &= x\,e^x - \int 1\cdot e^x\,dx \\ &= x\,e^x - e^x + C. \end{aligned}$$
   First function times integral of second ($x\cdot e^x$⁠), minus the integral of (derivative of first) times (integral of second), which is $\int 1\cdot e^x\,dx = e^x$⁠.
   $\int e^x\,dx = e^x + C$ (formula (xi)). Basics: integration undoes differentiation

3. Check: differentiate the answer

   $$\begin{aligned} \frac{d}{dx}\big(x e^x - e^x\big) &= (1\cdot e^x + x e^x) - e^x \\ &= x e^x. \end{aligned}$$
   The product rule on $x e^x$ gives $e^x + x e^x$⁠, and the $-e^x$ removes the extra $e^x$⁠. The derivative is exactly the integrand, for every real $x$⁠.
   Product rule; $\frac{d}{dx}e^x = e^x$⁠. Basics: integration undoes differentiation

Every fact used, and where it comes from