A __rationalizing substitution__ is a substitution that you make in order to turn an integrand into a _rational function._
Once you have a rational funtion, you can use __Integration by Partial Fraction Decomposition__ to integrate the function.
This technique wasn't taught in Calculus II (MAC2312 in Florida), but it can be helpful.
Whenever you see that your integrand would essentially be a rational function, if not for the presence of square roots, cube roots, or n-roots, you might want to try this technique.
But try U-Substitution first.
If there is an $x^2$ term under a square root (e.g. $sqrt{4-x^2}$), you should try using a Trigonometric Substitution to integrate it first, if appropriate.
Let $u$ be equal to the entire _n_th root term. Then, raise both sides by the _n_th power, so that there are no roots left.
Implicitly differentiate both sides with respect to $x$ (or $t$, or whatever your variable is).
Using algebra, isolate the $dx$ term and substitute.
For example, if you are given the following:
$$\int \frac{x^3}{\sqrt[3]{x^2+1}} dx$$You cannot use Trigonometric Substitution, because the root is a cube root; not a square root.
Consequently, you make a rationalizing substitution, by letting $u = \sqrt[3]{x^2+1}$
$$\implies u^3 = x^2+1$$Now, implicitly differentiate both sides with respect to $x$:
$$\implies 3u^2\frac{du}{dx} = 2x$$$$\implies dx = \frac{3u^2}{2x}du$$Substituting, you get:
$$= \int \frac{x^3}{u}\frac{3u^2}{2x}du$$$$= \frac{3}{2}\int x^2u du$$$$= \frac{3}{2} \int u^4 - u du$$$$= \frac{3}{2} \left( \frac{u^5}{5} - \frac{u^2}{2} \right) + C$$$$= \frac{3u^5}{10} - \frac{3}{4}u^2 + C$$
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