Integration by parts | Year 12

01 / LEARN

Understand the idea.

Reverse the product rule

Integrating the product rule produces $\int u d v = u v - \int v d u$ . Choose $u$ so differentiation simplifies it, and choose $d v$ so you can integrate it. Calculate $d u$ and $v$ explicitly before substituting into the formula.

A polynomial multiplied by an exponential or trigonometric function often becomes simpler when the polynomial is differentiated. For a lone logarithm, view the integrand as $ln x \cdot 1$ so that the easy-to-integrate factor is still available.

Repeat and verify

The remaining integral should be easier than the original. Repeated parts can reduce a polynomial degree one stage at a time. If an integral returns to itself, name it and collect its occurrences algebraically rather than repeating the same calculation indefinitely.

Keep the subtraction sign attached to the whole remaining integral. For definite integrals, either apply the limits to both terms of the formula or complete an antiderivative first. A derivative check is especially useful for finding missed product or chain factors.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

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

Worked solution

1
Choose and differentiate u.
$u = x, d v = e^{x} d x, d u = d x, v = e^{x}$
Differentiating x removes the polynomial factor while the exponential is easy to integrate.
2
Apply integration by parts.
$\int x e^{x} d x = x e^{x} - \int e^{x} d x$
The product-rule rearrangement subtracts the integral of v du.
3
Finish and check.
$\begin{array}{c} \int x e^{x} d x = (x - 1) e^{x} + C \\ \frac{d}{d x} [(x - 1) e^{x}] = e^{x} + (x - 1) e^{x} = x e^{x} \end{array}$
The two exponential terms combine to recover exactly the original product.

Answer

$(x - 1) e^{x} + C$ .

EXAMPLE 02

Worked example 2

Find $\int ln x d x$ for x>0.

Worked solution

1
Supply the implicit factor 1.
$u = ln x, d v = d x, d u = \frac{1}{x} d x, v = x$
The logarithm simplifies when differentiated, and its real domain here is x greater than zero.
2
Apply the product-rule formula.
$\int ln x d x = x ln x - \int x \cdot \frac{1}{x} d x$
The remaining product simplifies to 1 on the stated domain.
3
Integrate and verify.
$\begin{array}{c} \int ln x d x = x ln x - x + C, x > 0 \\ \frac{d}{d x} (x ln x - x) = ln x + 1 - 1 = ln x \end{array}$
Differentiation cancels the extra constant generated by the product rule.

Answer

$x ln x - x + C$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

For $\int x cos x d x$ , choose u and dv.

Show answer & reasoning

$u = x$ , $d v = cos x d x$ .

Choose $u = x$ and $d v = cos x d x$ . Then $d u = d x$ and $v = sin x$ , leaving $\int sin x d x$ , which is simpler than the original product.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find $\int x cos x d x$ .
2 marks
Find $\int x sin 2 x d x$ .
2 marks
Find $\int x^{2} e^{x} d x$ .
2 marks
Find $\int x ln x d x$ for $x > 0$ .
2 marks
Evaluate $\int_{0}^{1} x e^{x} d x$ .
2 marks
Find $\int arctan x d x$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $\int x cos x d x$ .
2 marks
Use $u = x$ , $v = sin x$ .
Subtract the integral of $sin x$ .
$x sin x + cos x + C$ .
Find $\int x sin 2 x d x$ .
2 marks
Use $v = - \frac{cos 2 x}{2}$ .
Integrate the remaining cosine term.
$- \frac{x cos 2 x}{2} + \frac{sin 2 x}{4} + C$ .
Find $\int x^{2} e^{x} d x$ .
2 marks
Apply parts with $u = x^{2}$ .
Apply parts again to the remaining $x e^{x}$ term.
$e^{x} (x^{2} - 2 x + 2) + C$ .
Find $\int x ln x d x$ for $x > 0$ .
2 marks
Use $u = ln x$ , $d v = x d x$ .
The remaining integrand is $\frac{x}{2}$ .
$\frac{x^{2}}{2} ln x - \frac{x^{2}}{4} + C$ .
Evaluate $\int_{0}^{1} x e^{x} d x$ .
2 marks
Use antiderivative $(x - 1) e^{x}$ .
The endpoint values are 0 and -1.
$1$ .
Find $\int arctan x d x$ .
2 marks
Use $u = arctan x$ , $d v = d x$ .
The remaining integral is $\frac{x}{1 + x^{2}}$ .
$x arctan x - \frac{1}{2} ln (1 + x^{2}) + C$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics integration techniques and applications, including inverse-trigonometric patterns, substitution, parts, simple partial fractions, areas and solids of revolution.

Official curriculum source ↗

In this lesson

Understand the idea.

Reverse the product rule

Repeat and verify

See the method in action.

Worked example 1

Choose and differentiate u.

Apply integration by parts.

Finish and check.

Worked example 2

Supply the implicit factor 1.

Apply the product-rule formula.

Integrate and verify.

Try it yourself.

Check your understanding

Keep practising