Evaluate an integral by integration by parts

Primary Tool

Math Workspace (CAS)

CAS is fitting here because it lets the page show the antiderivative, then immediately verify it by differentiating the answer.

Problem

Evaluate $\int x e^{x} d x$ .

This is a classic first example because one factor becomes simpler when differentiated and the other is unchanged when integrated.

The decision that matters

Taking $u = x$ and $d v = e^{x} d x$ means $d u = d x$ and $v = e^{x}$ . The integration-by-parts formula then turns the original problem into something simpler: $x e^{x} - \int e^{x} d x$ .

Checking the result by differentiation is not extra decoration. It confirms that the antiderivative matches the original integrand exactly.

Step-by-step walkthrough

This example is useful because every step is visible and there is only one remaining integral after applying the formula.

1Choose $u = x$ and $d v = e^{x} d x$ so that differentiation simplifies one factor and integration leaves the other unchanged.
2Compute $d u = d x$ and $v = e^{x}$ .
3Apply integration by parts: $\int u d v = uv - \int v d u$ , which gives $\int x e^{x} d x = x e^{x} - \int e^{x} d x$ .
4Integrate the remaining term to get $x e^{x} - e^{x} + C$ .
5Factor the common exponential if you want a cleaner final form: $e^{x} (x - 1) + C$ .
6Differentiate the answer to verify that it returns $x e^{x}$ .

Common Pitfall

Students sometimes think integration by parts always makes a problem easier. It only helps when your choice of $u$ actually simplifies the remaining integral.

Try a Variation

Try the same method on $\int x cos x d x$ . Which part of the workflow stays the same, and which antiderivative changes?

Keep moving through the cluster

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ExplanationCalculus2D Graphing

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The derivative is the slope of the tangent line, but that phrase only becomes useful when you compare the curve to nearby secant lines and local linear approximations.

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Worked ExampleCalculusMath Workspace (CAS)

Differentiate a polynomial from first principles

This worked example shows how the derivative definition creates an algebra problem first and a limit problem second. The simplification step is the point of the exercise.

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