# 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.

-
I keep seeing this here. People want to "solve" an integral. One solves an equation, or solves a problem. On evaluates an expression, including integrals. – Michael Hardy Sep 12 '11 at 1:29
@Michael Hardy: Math is not mine primary interest, so please excuse me on using inaproppriate terms, but this kind of tasks is very important to me right now, and after passing the upcoming exam, i don't plan to "solve" anything like this :D – jbennet Sep 12 '11 at 23:33

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.

-

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

$$\int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}dx = \frac{3 (4x - 1)^{2/3} \left(80 x^2 - 232 x + 2153\right)}{2560} + C$$

-