[justify]When the integrand is formed by a product (or a division, which we can treat like a product) it's recommended the use of the method known as [b]integration by u-substitution[/b], that consists in applying the following formula:[br][br][math]\int u·dv=u·v-\int v·du[/math][/justify]

[list]Even though it's a simple formula, it has to be applied correctly. Let's see a few tips on how to apply it well:[br][*][justify][b]Select [i]u[/i] and [i]dv[/i] correctly:[/b] A bad choice can complicate the integrand. Supposing we have a product, and one of the factors is monomial ([i]x[sup]3[/sup][/i] for example). If we consider that [i]dv = x[sup]3[/sup][/i], then by using integration we obtain that[br][math]v=\frac{x^4}{4}[/math][br]We have increased the exponent and this could mean a step back in the process.[br]Something similar happens with fractions (like [i]1/x[/i]). If we take [i]dv = 1/x[/i], we will obtain[i]v = log|x|[/i], and probably end up with a harder integration process.[br][/justify][justify]As a rule, we will call [i]u[/i] all powers and logarithms; and [i]dv[/i] exponentials, fractions and trigonometric functions (circular functions).[/justify][/*][*] [b]Don't change our minds about the selection:[/b] Sometimes we need to apply the method more than once for the same integral. When this happens, we need to call [i]u[/i] the result of [i]du[/i] from the first integral we applied the method to. The same applies to [i]dv[/i]. If we don't do this, seeing as choosing one option or another involves integration or differentiating, we'll be undoing the previous step and we won't be able to advance.[br][br][br][/*][*][justify][b]Cyclic integrals:[/b] Sometimes, after applying integration by u-substitution twice we have to isolate the very integral from the equality we've obtained in order to resolve it. [/justify][/*][/list]

[b]Example 1[br][/b][img]https://www.matesfacil.com/primitivas/IntPorPartes3.jpg[/img][br]In this integral we don't have an explicit product of functions, but we don't know what the logarithms primitive function is, so we differentiate it, that way [i]u = ln(x)[/i].[br][br][img]https://www.matesfacil.com/primitivas/IntPorPartes3-1.jpg[/img][br][br][b]Example 2[br][br][/b]It's in our interest to select [i]u = x[sup]2[/sup][/i] (to reduce the exponent) but then we're forced that [i]dv = ln(x)[/i] and obtaining [i]v[/i] isn't immediate. So we'll select the other case[br][br][img]https://www.matesfacil.com/primitivas/IntPorPartes4-1.jpg[/img][br][br][b]More examples:[/b] [url=][b]Integration by U-Substitution:[/b] resolved integrals step by step[/url]