Home  18.013A  Chapter 19 


The second useful tool is the backward version of the product rule. The product rule, as we have noted often, tells us
This means that if we seek an antiderivative of $h(x)$ and we can write $h$ as $fg\text{'}$ , then we can write $fg\text{'}$ as $(fg)\text{'}f\text{'}g$ , and an antiderivative of $fg\text{'}$ is then the difference between any antiderivative of $(fg)\text{'}$ and one of $f\text{'}g$ .
But an antiderivative of $(fg)\text{'}$ is given by $fg$ ; so we can use the product rule here to reduce the problem of finding an antiderivative of $fg\text{'}$ to finding an antiderivative of $f\text{'}g$ , for any $f$ and $g$ .
This tool is useful for finding antiderivatives of products of the form $A(x)x$ if you know an antiderivative $B(x)$ for $A(x)$ and an antiderivative $C(x)$ for $B(x)$ as well.
We can set $f=x$ and $g=B$ in the identity above, and write $A(x)x=B\text{'}(x)x$ , which by this procedure is $(Bx)\text{'}B$ , where we have used the identity $x\text{'}=1$ . This has $BxC$ as an antiderivative and $BxC$ therefore an antiderivative of $Ax$ .
The procedure is called integration by parts. It is useful for finding antiderivatives of products of exponentials and powers or of trigonometric functions and powers or of logarithms and powers, among other things.
For example, suppose we want to integrate $x\mathrm{ln}xdx$ , that is, we seek the antiderivative of $x\mathrm{ln}x$ with respect to $x$ .
If we set
$u=\mathrm{ln}x$
and
$dv=xdx$
, we can deduce that
$du=\frac{dx}{x}$
and
$v=\frac{{x}^{2}}{2}$
is a possible antiderivative of
$x$
.
Integrating by parts tells us then
$udv=(uv)\text{'}vdu$
which gives, after integrating
Exercises:
Try integrating the following integrands with respect to $x$ by using this technique:
19.1. ${x}^{4}(\mathrm{ln}x)$
19.2. $x\mathrm{sin}x$
19.3. $x\mathrm{exp}x$
19.4. $(\mathrm{sin}x)\mathrm{exp}x$ (Hint: integrate by parts twice and solve the resulting equation.)
19.5. $x(\mathrm{sin}x)\mathrm{exp}x$
