Integration by Parts

Reference > Calculus: Integration

Description

A method of integration that transforms the integral of a product of two functions using the following:

\(\int u \, dv=uv-\int v \, d u\)
The goal is to transform the integral into another form that is easier to solve. It is the inverse of the product rule in differentiation.


Examples

Example 1

\[\int x{e}^{x} \, dx\]
1
Use Integration by Parts on \(\int x{e}^{x} \, dx\).
Let \(u=x\), \(dv={e}^{x}\), \(du=dx\), \(v={e}^{x}\)

2
Substitute the above into \(uv-\int v \, du\).
\[x{e}^{x}-\int {e}^{x} \, dx\]

3
The integral of \({e}^{x}\) is \({e}^{x}\).
\[x{e}^{x}-{e}^{x}\]

4
Add constant.
\[x{e}^{x}-{e}^{x}+C\]

Done


 

Example 2

\[\int \frac{\ln{x}}{{x}^{5}} \, dx\]
1
Use Integration by Parts on \(\int \frac{\ln{x}}{{x}^{5}} \, dx\).
Let \(u=\ln{x}\), \(dv=\frac{1}{{x}^{5}}\), \(du=\frac{1}{x} \, dx\), \(v=-\frac{1}{4{x}^{4}}\)

2
Substitute the above into \(uv-\int v \, du\).
\[-\frac{\ln{x}}{4{x}^{4}}-\int -\frac{1}{4{x}^{5}} \, dx\]

3
Use Constant Factor Rule: \(\int cf(x) \, dx=c\int f(x) \, dx\).
\[-\frac{\ln{x}}{4{x}^{4}}+\frac{1}{4}\int \frac{1}{{x}^{5}} \, dx\]

4
Use Power Rule: \(\int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C\).
\[-\frac{\ln{x}}{4{x}^{4}}-\frac{1}{16{x}^{4}}\]

5
Add constant.
\[-\frac{\ln{x}}{4{x}^{4}}-\frac{1}{16{x}^{4}}+C\]

Done


 

Example 3

\[\int x\cos{(3x)} \, dx\]
1
Use Integration by Parts on \(\int x\cos{3x} \, dx\).
Let \(u=x\), \(dv=\cos{3x}\), \(du=dx\), \(v=\frac{\sin{3x}}{3}\)

2
Substitute the above into \(uv-\int v \, du\).
\[\frac{x\sin{3x}}{3}-\int \frac{\sin{3x}}{3} \, dx\]

3
Use Constant Factor Rule: \(\int cf(x) \, dx=c\int f(x) \, dx\).
\[\frac{x\sin{3x}}{3}-\frac{1}{3}\int \sin{3x} \, dx\]

4
Use Integration by Substitution on \(\int \sin{3x} \, dx\).
Let \(u=3x\), \(du=3 \, dx\), then \(dx=\frac{1}{3} \, du\)

5
Using \(u\) and \(du\) above, rewrite \(\int \sin{3x} \, dx\).
\[\int \frac{\sin{u}}{3} \, du\]

6
Use Constant Factor Rule: \(\int cf(x) \, dx=c\int f(x) \, dx\).
\[\frac{1}{3}\int \sin{u} \, du\]

7
Use Trigonometric Integration: the integral of \(\sin{u}\) is \(-\cos{u}\).
\[-\frac{\cos{u}}{3}\]

8
Substitute \(u=3x\) back into the original integral.
\[-\frac{\cos{3x}}{3}\]

9
Rewrite the integral with the completed substitution.
\[\frac{x\sin{3x}}{3}+\frac{\cos{3x}}{9}\]

10
Add constant.
\[\frac{x\sin{3x}}{3}+\frac{\cos{3x}}{9}+C\]

Done


 
See Also

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