Finding the integral of $x^2\sqrt[3]{1-x}$

Question

$$\int x^2\sqrt[3]{1-x}dx$$ I'm just starting out integration and found the above question on a textbook. It is meant to be worked out by substitution but even after a day of struggling, I can't get the answer given in the book. I tried substituting u = $\sqrt[3]{1-x}$ and then tried 1-x = u. On both the cases I am getting an answer but both are different from the answer given in the book. The answer given in the book is : $$-\frac{3}{140}\left(35-40x+14x^2 \right)\left( 1-x\right)^{4/3}$$ I'm getting a common factor -3 and $(1-x)^{4/3}$ but not 140 and the terms inside the other bracket are different.

Answer 1

With the substitution $u = 1-x$, $du = -dx$, we obtain $$\int x^2 (1-x)^{1/3} \, dx = -\int (1-u)^2 u^{1/3} \, du = -\int u^{7/3} - 2u^{4/3} + u^{1/3} \, du.$$ Integrating term by term, we get $$-\frac{3u^{10/3}}{10} + \frac{6u^{7/3}}{7} - \frac{3u^{4/3}}{4} + C.$$ Factoring out a common term of $-\frac{3}{140}u^{4/3}$ we get $$-\frac{3u^{4/3}}{140} (14u^2 - 40u + 35) + C.$$ Expressing this in terms of $x$ gives $$-\frac{3(1-x)^{4/3}}{140} (14x^2 + 12x + 9) + C.$$

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You should note that in the last step, expressing the antiderivative in terms of $x$ gives a different quadratic factor: perhaps this is the one you obtained? The answer that you claim your book has given is not correct.

Answer 2

setting $$t=(1-x)^\frac13$$ then we get $$x=1-t^3$$ and $$dx=-3t^2dt$$ and our integral is $$\int (1-t^3)^2\cdot t\cdot (-3t^2)dt$$ which is simplier.

Answer 3

It would be nicer if $x$ carried the exponent $\frac13$ and $(1-x)$ carried the exponent $2$, because you could then just expand the latter and multiply through by the former to produce a simple sum of terms.

So, let's write $u=1-x$ (the thing under the radical). Then we can also get the part outside of the radical in terms of $u$ as well, because then $x=1-u$ and $dx = -du$. So the integral can be written as $$\int \underbrace{(1-u)^2}_{x^2}\underbrace{u^{1/3}}_{\sqrt[3]{1-x}}\underbrace{(-1)\; du}_{dx}$$ $$=\int -(1-2u + u^2)u^{1/3}\; du$$ $$=-\int (u^{1/3}-2u^{4/3}+u^{7/3})\; du$$ $$=-(\tfrac34u^{4/3} - \tfrac67u^{7/3} + \tfrac3{10}u^{10/3})+C$$ $$=- \left( \tfrac{105}{140}-\tfrac{120}{140}u+\tfrac{42}{140}u^2\right)u^{4/3}+C$$ $$=\tfrac3{140}(35 +40 u + 14 u^2)u^{4/3}+C$$ $$=\tfrac3{140}[35 +40(1-x) + 14 (1-x)^2](1-x)^{4/3}+C$$ $$=\tfrac3{140}[35 -(40-40x) + (14-28x+14x^2)](1-x)^{4/3}+C$$ $$=\boxed{\tfrac3{140}(9 -12 x + 14x^2)(1-x)^{4/3}+C}$$