# Integration by Parts

## Reference > Calculus: Integration

 DescriptionA 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 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$Donex*e^x-e^x+C  Example 2 [Top]$\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$Done-ln(x)/(4*x^4)-1/(16*x^4)+C  Example 3 [Top]$\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 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(x*sin(3*x))/3+cos(3*x)/9+C