Math · Calculus II · Worked example

Integrate by parts twice

Find the integral of x²eˣ.

\int x^{2} e^{x} d x

First round

Take u = x² and dv = eˣ dx, so du = 2x dx and v = eˣ.

\int x^{2} e^{x} d x = x^{2} e^{x} - \int 2 x e^{x} d x

Second round

The new integral is twice the integral of xeˣ, which the first worked example found: xeˣ − eˣ.

\int 2 x e^{x} d x = 2 x e^{x} - 2 e^{x} + C

Combine

Subtract the second result from x²eˣ and factor out eˣ.

\int x^{2} e^{x} d x = e^{x} (x^{2} - 2 x + 2) + C

Check by differentiating

The derivative of eˣ(x² − 2x + 2) is eˣ(x² − 2x + 2) + eˣ(2x − 2) = x²eˣ.

Result

eˣ(x² − 2x + 2) + C.

Your turn

Find the integral of eˣ sin x.

Show the answer and explanation

eˣ(sin x − cos x)/2 + C.

Call the integral I. Parts with u = sin x gives I = eˣ sin x − ∫eˣ cos x dx, and parts again with u = cos x gives I = eˣ sin x − eˣ cos x − I. So 2I = eˣ(sin x − cos x).

\begin{array}{l} \int e^{x} sin x d x \\ \frac{e^{x} (sin x - cos x)}{2} + C \end{array}

Keep exploring

In Derivative & antiderivative checks, enter eˣ(x² − 2x − 2), with one sign slipped. It does not agree: its derivative is eˣ(x² − 4).

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OpenStax Calculus Volume 2 — Integration by parts

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