Learning Lab

Formula

The integration by parts formula is obtained from the product rule for differentiation. For any two differentiable functions \(v\left(x\right)\) and \(u\left(x\right)\) we have, by the product rule, \[\begin{align*} \frac{d}{dx}\left(v\left(x\right)u\left(x\right)\right) & =v\left(x\right)\frac{du\left(x\right)}{dx}+u\left(x\right)\frac{dv\left(x\right)}{dx}. \end{align*}\] Integrating both sides with respect to \(x\) gives \[\begin{align*} \int\frac{d}{dx}\left(v\left(x\right)u\left(x\right)\right)dx & =\int\left(v\left(x\right)\frac{du\left(x\right)}{dx}+u\left(x\right)\frac{dv\left(x\right)}{dx}\right)dx,\;\textrm{that is,}\\ v\left(x\right)u\left(x\right) & =\int v\left(x\right)\frac{du\left(x\right)}{dx}dx+\int u\left(x\right)\frac{dv\left(x\right)}{dx}dx. \end{align*}\] This can be rearranged to give the formula for integration by parts: \[\begin{align*} \int v\left(x\right)\frac{du\left(x\right)}{dx}dx & =v\left(x\right)u\left(x\right)-\int u\left(x\right)\frac{dv\left(x\right)}{dx}dx. \end{align*}\] Which is often abbreviated to \[\begin{align*} \int vu'dx & =vu-\int uv'dx. \end{align*}\]

The hope is that by choosing the functions \(v\) and \(u'\) carefully, the integral on the right hand side is easier than the original integral. Successful use of integration by parts requires a correct choice for \(v\) and \(u'\). If the integral on the right is harder, then try swapping \(v\) and \(u'\).

To assist you in choosing \(v\) and \(u'\) choose \(v\) to be the function that comes first in the sequence of functions:

1. logarithmic function

2. inverse trig function

3. algebraic function

4. trigonometric function

5. exponential function

\(u'\) is then the other function in the integrand. 11 The integrand is the function that is being integrated. For example the integrand in \[ \int x\sin\left(x\right)dx \] is \(x\sin\left(x\right)\). In this case it is the product of an algebraic function \(x\) and a trigonometric function \(\sin\left(x\right)\) so we would set \(v=x\), (because algebraic functions are higher on the list than trigonometric functions), and \(u'=\sin\left(x\right)\).

Repeated Use of Integration by Parts

Some integrals require the use of the integration by parts technique more than once. For example, consider \(\int x^{2}e^{x}dx\). We have an algebraic function \(x^{2}\) multiplying an exponential function \(e^{x}.\) According to our guide we set \(v\left(x\right)=x^{2}\) and \(u'\left(x\right)=e^{x}.\) Then \(dv/dx=2x\) and \(u\left(x\right)=e^{x}.\) Substituting in the integration by parts formula we get: 88 We use \[\begin{align*} \int vu'dx & =vu-\int uv'dx. \end{align*}\] \[\begin{align*} \int x^{2}e^{x}dx & =x^{2}e^{x}-\int e^{x}\left(2x\right)dx. & (2) \end{align*}\] The integral on the right hand side requires another integration by parts. We set \(v\left(x\right)=2x\) and \(u'\left(x\right)=e^{x}.\) Then \(dv/dx=2\) and \(u\left(x\right)=e^{x}.\) Substituting in the integration by parts formula gives \[\begin{align*} \int e^{x}\left(2x\right)dx & =2xe^{x}-\int2e^{x}dx\\ & =2xe^{x}-2e^{x}+c,\;c\in\mathbb{R}. & (3) \end{align*}\] Now substitute this result in equation (2) to get the result \[\begin{align*} \int x^{2}e^{x}dx & =x^{2}e^{x}-\int e^{x}\left(2x\right)dx\\ & =x^{2}e^{x}-\left(2xe^{x}-2e^{x}+c\right)\;\textrm{using ($3$ )}\\ & =x^{2}e^{x}-2xe^{x}+2e^{x}-c\\ & =e^{x}\left(x^{2}-2x+2\right)-c. \end{align*}\] Note that it doesn’t matter if we put \(c\) or \(-c\) as \(c\) is just any constant.

Exercises

\[\begin{array}{lll} \textrm{1. Find}\\ a)\;\int x\sin\left(x\right)dx & & b)\;\int xe^{x}dx\\ c)\;\int x\cos\left(4x\right)dx & & d)\;\int x^{4}\ln\left(x\right)dx\\ \\ \textrm{2. Find}\\ a)\;\int x^{2}\cos\left(x\right)dx & & b)\;\int x^{3}e^{x^{2}}\;\textrm{(Hint: let $v=x^{2})$ } \end{array}\]