Integration by parts

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Integration by parts

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 of inverse of the differential product rule. Say u and v are functions of x, then we have that

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

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

Integrating both sides gives us

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

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

This means that we can transform one integral into 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

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