Integration by Parts

Integration by parts is used to convert a difficult integral into an easier one. It comes from reversing the product rule for differentiation.

The rule is:

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

Steps

Identify \(u\) and \(\frac{dv}{dx}\).
Find \(\frac{du}{dx}\) and \(v\).
Substitute into the formula.

Go again if necessary !

Tip: Choose \(u\) as the function that becomes simpler when differentiated.

Example

Find the integral:

\[ \int x\sin x \, dx \]

Choose:

\[ u = x, \qquad \frac{dv}{dx} = \sin x\ \]

Then:

\[ \frac{du}{dx} = 1 \quad\text{and}\quad v = \int \sin x \, dx = -\cos x. \]

Apply the formula:

\[ \int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx \] \[ \int x\sin x\,dx = x(-\cos x) - \int (-\cos x)\,dx \] \[ \int x\sin x\,dx = -x\cos x + \int \cos x\,dx \] \[ \int x\sin x\,dx = -x\cos x + \sin x + C \]

Example

Find the integral:

\[ \int x\ln x\,dx \]

Choose:

\[ u = \ln x, \qquad \frac{dv}{dx} = x. \]

Then:

\[ \frac{du}{dx} = \frac{1}{x}, \qquad v = \frac{x^2}{2}. \]

Apply the formula:

\[ \int x\ln x\,dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\frac{1}{x}\,dx \] \[ = \frac{x^2}{2}\ln x - \frac{1}{2}\int x\,dx \] \[ = \frac{x^2}{2}\ln x - \frac{1}{2}\cdot\frac{x^2}{2} + C \] \[ = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C. \]

Example

Find the integral:

\[ \int x^2 e^x\,dx \]

This requires integration by parts twice.

First pass:

\[ u = x^2, \qquad \frac{dv}{dx} = e^x \] \[ \frac{du}{dx} = 2x, \qquad v = e^x \]

\[ \int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx \] \[ \int x^2 e^x\,dx = x^2 e^x - \int e^x(2x)\,dx \] \[ \int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx \]

Go again:

\[ \int 2x e^x\,dx \] \[ \text{Let } u = 2x \quad\text{and}\quad \frac{dv}{dx} = e^x \] \[ \frac{du}{dx} = 2 \quad\text{and}\quad v = e^x \]

\[ \int 2x e^x\,dx = 2x e^x - \int 2 e^x\,dx \] \[ \text{So } \int x^2 e^x\,dx = x^2 e^x - \left( 2x e^x - \int 2 e^x\,dx \right) \] \[ \int x^2 e^x\,dx = x^2 e^x - 2x e^x + 2\int e^x\,dx \] \[ \int x^2 e^x\,dx = x^2 e^x - 2x e^x + 2e^x + C \] \[ \int x^2 e^x\,dx = e^x(x^2 - 2x + 2) + C \]