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 [Top]

\[\int x{e}^{x} \, dx\]
1
Apply Integration By Parts to \(\int x{e}^{x} \, dx\)
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 [Top]

\[\int \frac{\ln{x}}{{x}^{5}} \, dx\]
1
Apply Integration By Parts to \(\int \frac{\ln{x}}{{x}^{5}} \, dx\)
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
Apply the 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
Apply the 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 [Top]

\[\int x\cos{(3x)} \, dx\]
1
Apply Integration By Parts to \(\int x\cos{3x} \, dx\)
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
Apply the 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
Apply Integration By Substitution to \(\int \sin{3x} \, dx\)
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
Apply the Constant Factor Rule: \(\int cf(x) \, dx=c\int f(x) \, dx\)
\[\frac{1}{3}\int \sin{u} \, du\]

7
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