I = (x+1)*(1-x)^(1/3) dx
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Originally Posted by jdesak I = (x+1)*(1-x)^(1/3) dx Write it as $\displaystyle \int {{u^{\frac{1}{3}}}} \left( {2 - u} \right)\left( { - du} \right)$ and expand.
In other words, because it is that "1/3 power" that is the hard part, use the substitution u= 1- x (or equivaently u= x- 1).
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