Editing Integration by substitution (section)

=== Introduction (indefinite integrals) ===
Before stating the result [[mathematical rigor|rigorously]], consider a simple case using [[indefinite integral|indefinite integrals]].

Compute <math display="inline">\int(2x^3+1)^7(x^2)\,dx.</math><ref>{{harvnb|Swokowski|1983|p=258}}</ref>

Set <math>u=2x^3+1.</math> This means <math display="inline">\frac{du}{dx}=6x^2,</math> or as a [[differential form]], <math display="inline">du=6x^2\,dx.</math> Now: 
<math display="block">\begin{aligned}
    \int(2x^3 +1)^7(x^2)\,dx
    &= \frac{1}{6}\int\underbrace{(2x^3+1)^{7}}_{u^{7}}\underbrace{(6x^2)\,dx}_{du} \\
    &= \frac{1}{6}\int u^{7}\,du \\
    &= \frac{1}{6}\left(\frac{1}{8}u^{8}\right)+C \\
    &= \frac{1}{48}(2x^3+1)^{8}+C,
\end{aligned}</math>
where <math>C</math> is an arbitrary [[constant of integration]].

This procedure is frequently used, but not all integrals are of a form that permits its use. In any event, the result should be verified by differentiating and comparing to the original integrand.
<math display="block">\frac{d}{dx}\left[\frac{1}{48}(2x^3+1)^{8}+C\right] = \frac{1}{6}(2x^3+1)^{7}(6x^2) = (2x^3+1)^7(x^2).</math>
For definite integrals, the limits of integration must also be adjusted, but the procedure is mostly the same.