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For example, how would I know to make my $u$ the $\sqrt{1-x}$ when given a problem such as to evaluate the integral of $x^2\sqrt{1-x}$? Is there a logical thinking you use to know $u = \sqrt{1-x}$ rather than just $1-x$? I just learned $u$-substitution the other day, but my teacher stressed that in nearly all cases we will not have a u raised to a power.

Any info on this would be grateful!

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    Thanks Joe Johnson 126 for editing my functions in Latex.2012-02-08
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    You can also use $u=1-x$. Then $du=-dx$ and $x^2=(1-u)^2$. This leads to the integral $$\int -(1-u)^2\sqrt{u}\,du$$. Expand the $(1-u)^2$ and distribute the $\sqrt{u}$. Use power rule.2012-02-08
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    Radicals are always tough to deal with; thus, one tries to find a substitution that gets rid of them. In particular, if we look at the factor $u=\sqrt{1-x}$, we can try inverting this relation to yield $x=1-u^2$, $\mathrm dx=-2u \mathrm du$, which is in fact a perfectly working substitution...2012-02-08
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    @user24641: No problem. I love to TeX.2012-02-09
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    I somewhat prefer $1-x=u^2$.2012-02-09

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$u=\sqrt{1-x}$ is an example of a "rationalizing substitution". "Rationalizing" means getting rid of the radical. One writes $u^2=1-x$, so $2u\;du = -dx$. Then $x$ is replaced by $1-u^2$. The point is that putting the problem in a form where there are no other functions besides polynomials enables one to rely on the fact that one knows how to deal with polynomials.