# Integration by parts

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In the case the integrand is a product of two functions we may be able to solve it by substitution, as shown in the previous pages. But what if does not work? One way one could try is a kind if inverse of the differential product rule. Say u and v are functions of x, then we have that $\frac{d}{{dx}}\left( {uv} \right) = \frac{{du}}{{dx}}v + u\frac{{dv}}{{dx}}$

or $u\frac{{dv}}{{dx}} = \frac{d}{{dx}}\left( {uv} \right) - \frac{{du}}{{dx}}v$

Integrating both sides gives us $\int {u\frac{{dv}}{{dx}}dx} = \int {\frac{d}{{dx}}\left( {uv} \right)} - \int {\frac{{du}}{{dx}}v\;dx}$

or $\int {u\frac{{dv}}{{dx}}dx} = uv - \int {\frac{{du}}{{dx}}v\;dx}$

This means that we can transform one integral to another that may be solvable. The goal is thus to find $\int {f\left( x \right)g\left( x \right)dx}$

Example: First an example to make some sense of this. Let us try to find $\int {x{e^x}} dx$

We cannot use the substitution method since none of the inner derivatives of the other function is the derivative of the other function. But let us instead set $\left| {\begin{array}{*{20}{c}} {u = x} \\ {\frac{{dv}}{{dx}} = {e^x}}\end{array}} \right.$

and thus $\left| {\begin{array}{*{20}{c}} {\frac{{du}}{{dx}} = 1} \\ {v = \int {{e^x}dx} = {e^x}}\end{array}} \right.$

That means that $\begin{gathered} \int {\underbrace x_u\underbrace {{e^x}}_{dv/dx}} dx = \underbrace x_u\underbrace {{e^x}}_v - \int {\underbrace 1_{du/dx}} \underbrace {{e^x}}_vdx \hfill \\ \quad = x{e^x} - {e^x} + C = {e^x}(x - 1) + C \hfill \\ \end{gathered}$

We can test this by differentiating: $\frac{d}{{dx}}({e^x}(x - 1) + C) = {e^x}(x - 1) + {e^x}1 = {e^x}x - {e^x} + {e^x} = x{e^x}$

Other ways to write the rule

You might find this rule written as $\int {fg\;dx} = Fg - \int {Fg'\;dx}$

Here the two functions have swapped order.  Using this way of writing it might be a shortcut. $\int {\underbrace x_g\underbrace {{e^x}}_{\text{f}}} dx = {e^x}x - \int {{e^x} \cdot 1\;dx} = {e^x}x + {e^x} + C$

You might also see it written as $\int {u\;dv} = uv - \int {v\;du}$

Here we have that $dv = \frac{{dv}}{{dx}}dx$

In the next few pages we will look at some cases. Up a level : Integrals
Previous page : Integration of definite integrals using substitution
Next page : One function is in the form x to the power of n and the other can be integrated at least n times Last modified: May 20, 2018 @ 13:45