3. Integration by Parts

Video coming soon5 worked examples

Theory

Integration by parts reverses the product rule:

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx

Choose $u$ to be the factor that becomes simpler when differentiated. A useful guide is LIATE (Logarithm, Inverse trig, Algebraic, Trig, Exponential): the factor earliest in this list is usually $u$ . Some integrals need parts applied more than once.

The Golden Rule: pick $u$ so that $\frac{du}{dx}$ is simpler, and $\frac{dv}{dx}$ as the part you can integrate; then apply $uv - \int v\,\frac{du}{dx}\,dx$ .

⚠️ Common Examiner Traps

Wrong choice of $u$ : if the new integral is harder, you likely chose $u$ and $dv$ the wrong way round.
Sign error: the formula subtracts the second integral.
The $\ln x$ trick: for $\int \ln x\,dx$ , take $u = \ln x$ and $\frac{dv}{dx} = 1$ .

Worked examples

Example 1

Find $\displaystyle\int x e^x\,dx$ .

Step 1: Let $u = x$ (so $\frac{du}{dx} = 1$ ) and $\frac{dv}{dx} = e^x$ (so $v = e^x$ ).

Step 2: Apply the formula:

\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + c

Step 3: Factor: $= e^x(x - 1) + c$ .

Example 2

Find $\displaystyle\int x \cos x\,dx$ .

Step 1: Let $u = x$ and $\frac{dv}{dx} = \cos x$ (so $v = \sin x$ ).

Step 2: Apply the formula:

\int x \cos x\,dx = x \sin x - \int \sin x\,dx = x \sin x + \cos x + c

Example 3

Find $\displaystyle\int \ln x\,dx$ .

Step 1: Take $u = \ln x$ (so $\frac{du}{dx} = \frac{1}{x}$ ) and $\frac{dv}{dx} = 1$ (so $v = x$ ).

Step 2: Apply the formula:

\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - \int 1\,dx

Step 3: So $\int \ln x\,dx = x\ln x - x + c$ .

Example 4

Find $\displaystyle\int x^2 e^{3x}\,dx$ .

Step 1: Take $u = x^2$ so that differentiating reduces the power. With $\frac{dv}{dx} = e^{3x}$ , we get $v = \frac{1}{3}e^{3x}$ :

\int x^2 e^{3x}\,dx = \frac{1}{3}x^2e^{3x} - \frac{2}{3}\int x e^{3x}\,dx

Step 2: The remaining integral still has a power of $x$ , so apply integration by parts a second time, now with $u = x$ :

\int x e^{3x}\,dx = \frac{1}{3}xe^{3x} - \frac{1}{3}\int e^{3x}\,dx = \frac{1}{3}xe^{3x} - \frac{1}{9}e^{3x}

Step 3: Substitute this back into Step 1, taking care with the $-\frac{2}{3}$ multiplying both terms:

\int x^2 e^{3x}\,dx = \frac{1}{3}x^2e^{3x} - \frac{2}{3}\left(\frac{1}{3}xe^{3x} - \frac{1}{9}e^{3x}\right)

Step 4: Expand and tidy:

= \frac{1}{3}x^2e^{3x} - \frac{2}{9}xe^{3x} + \frac{2}{27}e^{3x} + c

Each application of the rule drops the power of $x$ by one, so an $x^n$ term needs $n$ applications.

Example 5

Find $\displaystyle\int e^{2x}\sin x\,dx$ .

Step 1: Neither factor simplifies when differentiated, so parts will not terminate. Give the integral a name so we can treat it as an unknown:

I = \int e^{2x}\sin x\,dx

Step 2: Apply parts with $u = e^{2x}$ and $\frac{dv}{dx} = \sin x$ , so $v = -\cos x$ :

I = -e^{2x}\cos x + 2\int e^{2x}\cos x\,dx

Step 3: Apply parts again to the new integral, keeping $u = e^{2x}$ — switching now would simply undo the first step:

\int e^{2x}\cos x\,dx = e^{2x}\sin x - 2\int e^{2x}\sin x\,dx = e^{2x}\sin x - 2I

Step 4: The original integral has reappeared. Substitute back:

I = -e^{2x}\cos x + 2\left(e^{2x}\sin x - 2I\right) = -e^{2x}\cos x + 2e^{2x}\sin x - 4I

Step 5: Now solve for $I$ algebraically, gathering both $I$ terms on the left:

5I = e^{2x}\left(2\sin x - \cos x\right)

Step 6: Divide, remembering the constant of integration only at the end:

I = \frac{1}{5}e^{2x}\left(2\sin x - \cos x\right) + c