8. Integration by Parts

8.1. Integration by Parts

We start this section by thinking about differentiation rather than integration, and particularly the product rule which states that

\[\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{du}{dx}\]

Since both sides are equal, we expect the integral of both sides to also be equal (apart from the constants of integration) and so we can write

\[\int\frac{d}{dx}\left(uv\right)dx=\int u\left(\frac{dv}{dx}\right)dx+\int v\left(\frac{du}{dx}\right)dx+B\]

where B is a constant.

However we can see that \(\int\frac{d}{dx}\left(uv\right)dx=uv\) and so by rearrangement we have

\[\int u\left(\frac{dv}{dx}\right)dx=uv-\int v\left(\frac{du}{dx}\right)dx+c\]

although it is easier to write this in short-hand form

\[\int uv'=uv-\int vu'+c\]

It is useful to go through an example of integration by parts so that you can see the steps and how the process is best laid out.

For the integral

\[\int xe^{x}dx\]

we start by assigning u and v. Part of the skill of integrating by parts is assigning the right functions to u and v, otherwise you end up making the integrand more complicated. Let

\[\begin{split}\begin{aligned} u=x & & u'=1\\ v'=e^{x} & & v=e^{x}\end{aligned}\end{split}\]

and so we use the integration by parts formula to get

\[\begin{split}\begin{aligned} \int xe^{x}dx & = & xe^{x}-\int e^{x}dx+c\\ & = & xe^{x}-e^{x}+x\\ & = & e^{x}\left(x-1\right)+c\end{aligned} \end{split}\]

8.2. Integration using Partial Fractions

The method of partial fractions can be used to simplify some integrals. Lets take an example fraction that we want to integrate

\[\frac{\left(3x-1\right)}{\left(x-2\right)\left(x+3\right)}\]

which we can express as the sum of two partial fractions

\[\frac{\left(3x-1\right)}{\left(x-2\right)\left(x+3\right)}=\frac{A}{\left(x-2\right)}+\frac{B}{\left(x+3\right)}\]

Cross-multiplication and simplification gives

\[\begin{split}\begin{aligned} \left(3x-1\right) & = & A\left(x+3\right)+B\left(x-2\right)\\ \left(3x-1\right) & = & x\left(A+B\right)+\left(3A-2B\right)\end{aligned}\end{split}\]

Equating coefficients (if you have forgotten how to do this, look back at unit 1 notes) gives \(A=1\) and \(B=2\) , and so we can now rewrite the integral of the original function as

\[\begin{split}\begin{aligned} \int\frac{\left(3x-1\right)}{\left(x-2\right)\left(x+3\right)} & = & \int\frac{1}{\left(x-2\right)}dx+\int\frac{2}{\left(x+3\right)}dx\\ & = & \ln\left|x-2\right|+2\ln\left|x+3\right|+c\end{aligned} \end{split}\]

8.3. Logarithmic integration

You should already be familiar with the integral

\[\int\frac{1}{x}\,\mathrm{d}x=\ln\left|x\right|+c\]

but this is in fact a specific case of the general result

\[\int\frac{f'\left(x\right)}{f\left(x\right)}\,\mathrm{d}x=\ln\left|f\left(x\right)\right|+c\]

So, if the numerator is (or is related to) the first derivative of the denominator then the result is of the form \(\ln\left|f\left(x\right)\right|\).