Evaluate $\int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}\mathrm  dx$ I need help to evaluate the integral: $$ \int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}\mathrm dx.$$ 
The procedure on Wolfram Alpha is very long and complicated. Is there any easier way ?
Thanks.
 A: One way would start with a rationalizing substitution:
$$
\begin{align}
u & = \sqrt[3]{4x-1} \\
u^3 & = 4x-1 \\
3u^2\;du & = 4\;dx \\
\frac{u^3+1}{4} & = x
\end{align}
$$
So you have
$$
\int \frac{\left(\frac{u^3+1}{4}\right)^2 -2\left(\frac{u^3+1}{4}\right)+7}{u} \; \frac{3u^2\;du}{4}
$$
A $u$ in the numerator cancels out the denominator and then you're just integrating a polynomial.
A: As suggested in the comments, you can substitute to get rid of the $-1$ in the third root. Let $y = x - \frac{1}{4}$. To write $p(x) = x^2 - 2x + 7$ as a polynomial in $y$, note that $p(x) = p(y + \frac{1}{4})$ so substituting $x = y + \frac{1}{4}$ in $p(x)$ gives $x^2 - 2x + 7 = y^2 - \frac{3}{2} y + \frac{105}{16}$. So we get
$$\int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}dx = \int \frac{y^{2}-\frac{3}{2}y+\frac{105}{16}}{\sqrt[3]{4y}}dy = \frac{1}{\sqrt[3]{4}} \left( \int y^{5/3} dy - \frac{3}{2} \int y^{2/3} dy + \frac{105}{16} \int y^{-1/3} dy \right)$$
Solving each integral individually gives
$$\frac{1}{\sqrt[3]{4}} \left(\frac{3}{8} y^{8/3} - \frac{9}{10} y^{5/3} + \frac{315}{32} y^{2/3} + C\right)$$
Plugging in $y = x - \frac{1}{4}$ gives
$$\int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}dx = \frac{3 (4x - 1)^{2/3} \left(80 x^2 - 232 x + 2153\right)}{2560} + C$$
