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$$\int \frac{x^3}{(x^2 + 1)^\frac{1}{3}}dx$$

I am suppose to make a $u$ substitution and to make this a rational integral and then evaluate it from there but I have no idea how to do that. There aren't any good examples of this in the book and I can not find any us that make this doable.

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2 Answers 2

up vote 7 down vote accepted

Let $u = x^2 + 1$, $du = 2x\,dx$:

\begin{align*} \int \frac{x^3\,dx}{(x^2 + 1)^\frac{1}{3}} &= \int \frac{x^2\,x\,dx}{(x^2 + 1)^\frac{1}{3}} = \frac{1}{2} \int \frac{u-1}{u^\frac{1}{3}}\,du = \frac{1}{2} \int \left(u^\frac{2}{3} - u^{-\frac{1}{3}}\right) \,du \\ &= \frac{3}{10} u^{\frac{5}{3}} - \frac{3}{4} u^{\frac{2}{3}} + C \\ &= \frac{3}{5} \left(x^2+1\right)^{\frac{5}{3}} - \frac{3}{4} \left(x^2+1\right)^{\frac{2}{3}} + C \end{align*}

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Let $x^2 + 1 = t^3$. Hence, we get that $2x dx = 3t^2dt$ and $xdx = \dfrac{3t^2}{2} dt$. Hence, we have that \begin{align} I & = \int \dfrac{x^3}{(x^2+1)^{1/3}} dx = \int \dfrac{x^2 (xdx)}{(x^2+1)^{1/3}}\\ & = \int \dfrac{t^3-1}{t} \dfrac{3t^2}{2} dt = \dfrac32 \int(t^4-t) dt\\ & = \dfrac32 \left(\dfrac{t^5}{5} - \dfrac{t^2}{2} \right) + C\\ & = \dfrac32 \left( \dfrac{(x^2+1)^{5/3}}{5} - \dfrac{(x^2+1)^{2/3}}{2}\right) + C \end{align}

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