3.5 Integration by Parts
3.5 Integration by Parts
Integration by Parts
Integration by parts is an integration method which enables us to find antiderivatives of some new functions such as [latex]\ln(x)[/latex] as well as antiderivatives of products of functions such as [latex]x^2\ln(x)[/latex] and [latex]xe^x[/latex].
If the function we’re trying to integrate can be written as a product of two functions, [latex]u[/latex], and [latex]dv[/latex], then integration by parts lets us trade out a complicated integral for hopefully simpler one.
Integration by Parts Formula
For indefinite integrals:
[latex]\int u\, dv = uv-\int v\, du[/latex]
For definite integrals:
[latex]\int_a^b u\, dv = \left.uv\right|_a^b-\int_a^b v\, du[/latex]
Note that integration by parts is a modified version of undoing product rule.
The hardest part in integration by parts is picking the right parts of the integrand to represent [latex]u[/latex] and [latex]dv[/latex]. A general rule of thumb is to chose [latex]dv[/latex] to be the part that we can easily integrate and [latex]u[/latex] such that it produces a simpler function when differentiated.
Example 1
Integrate [latex]\int xe^x\, dx[/latex].
Answer: [latex]\int xe^x\, dx=?[/latex]
Note that this is a product of two functions, [latex]x[/latex] and [latex]e^x[/latex], and so basic rules don’t apply. We also notice there is no composition of functions and so integration by substitution also does not apply. We can try the integration by parts method.
We have two options for [latex]u[/latex] and [latex]dv[/latex]:
[latex]u=x,\ dv=e^x, dx[/latex]
and
[latex]u=e^x,\ dv=x\,dx[/latex]
In the first option, the integral of [latex]dv[/latex] is easily determined and the derivative of [latex]u[/latex] results in a simpler function (a constant).
In the second option, though the integral of [latex]dv[/latex] is easily determined, the derivative of [latex]u[/latex] does not result in a simpler function (it is the same function)
Therefore, the first option is the preferred one and so we break apart the product into two parts: [latex]u=x[/latex] and [latex]dv=e^x\, dx.[/latex]
We now calculate [latex]du[/latex], the derivative of [latex]u[/latex], and [latex]v[/latex], the integral of [latex]dv[/latex]:
[latex]u=x\Rightarrow \frac{du}{dx}=1\Rightarrow du=dx[/latex]
[latex]dv =e^x\,dx\Rightarrow v= \int e^x\, dx = e^x[/latex]
Using the Integration by Parts formula,
[latex]\int xe^x\, dx=uv-\int v\, du = xe^x - \int e^x\, dx=xe^x-e^x+C[/latex]
In the last example we could have chosen either [latex]x[/latex] or [latex]e^x[/latex] as our [latex]u[/latex], but had we chosen [latex]u=e^x[/latex], the second integral would have become messier, rather than simpler.
Example 2
Integrate [latex]\int\limits_1^4\, 6x^2\ln(x)\, dx[/latex].
Answer: [latex]\int\limits_1^4\, 6x^2\ln(x)\, dx[/latex]
Note that this is a product of two functions, [latex]6x^2[/latex] and [latex]\ln(x)[/latex], and so basic rules don’t apply. We also notice there is no composition of functions and so integration by substitution also does not apply. We can try the integration by parts method.
We have two options for [latex]u[/latex] and [latex]dv[/latex]:
[latex]u=6x^2,\ dv=\ln(x), dx[/latex]
and
[latex]u=\ln(x),\ dv=6x^2\,dx[/latex]
In the first option, there is no formula we know of yet for the integral of [latex]dv[/latex].
In the second option the integral of [latex]dv[/latex] is easily determined and the derivative of [latex]u[/latex] results in a simpler function (a power function).
Therefore, the second option is the preferred one and so we break apart the product into two parts: [latex]u=\ln(x)[/latex] and [latex]dv=6x^2\,dx.[/latex]
We now calculate [latex]du[/latex], the derivative of [latex]u[/latex], and [latex]v[/latex], the integral of [latex]dv[/latex]:
[latex]u=\ln(x)\Rightarrow \frac{du}{dx}=\frac{1}{x}\Rightarrow du=\frac{1}{x}\,dx[/latex]
[latex]dv =6x^2\,dx\Rightarrow v= \int 6x^2\, dx =2x^3[/latex]
Using the Integration by Parts formula,
[latex]\begin{align*} \int\limits_1^4 6x^2\ln(x)\, dx&=uv-\int v\, du \\ &=\left.\ln(x)\cdot \left(2x^3\right)\right|_1^4 - \int\limits_1^4 2x^3\left(\frac{1}{x}\,dx\right)\\ &=\left.2x^3\ln(x)\right|_1^4 - \int_1^4 2x^2\,dx\\ &=\left.2x^3\ln(x)\right|_1^4-\left.\frac{2}{3}x^3\right|_1^4\\ &=2(4)^3\ln(4)-2(1)^3\ln(1)-\left(\frac{2}{3}\cdot 4^3-\frac{2}{3}\cdot(1)^3\right)\\ &=128\ln(4)-\frac{128}{3}+\frac{2}{3}\\ &=128\ln(4)-42 \end{align*}[/latex]