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I've been given the following integral to evaluate

$$\int_0^1(3x^3 - x^2 + 2x - 4)\frac{1}{\sqrt{x^2 - 3x + 2}} dx.$$

Can I expand fractional powers?

thesmallprint
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Zoe
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  • Have a look at https://math.stackexchange.com/questions/2814179/how-to-integrate-the-product-of-two-or-more-polynomials-raised-to-some-powers-n/2814254#2814254 – Claude Leibovici Jun 11 '18 at 04:30

1 Answers1

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You should complete the square to obtain

$$ x^2-3x+2=\left(x-\frac{3}{2}\right)^2-\frac{1}{4} $$

then let $x-\frac{3}{2}=\frac{1}{2}\sec(t)$, that is, $x=\frac{3}{2}+\frac{1}{2}\sec(t)$. Then $dx=\frac{1}{2}\sec(t)\tan(t)\,dt$ and

$$ \sqrt{x^2-3x+2}=\pm\frac{1}{2}\tan(t) $$

For $0\le x\le1$, $\sec(t)<0$ and $\sec^{-1}(-3)\le t\le\pi$, so you will need

$$ \sqrt{x^2-3x+2}=-\frac{1}{2}\tan(t) $$

Upon substitution into the original, you obtain an integral whose terms are powers of the secant function.

John Wayland Bales
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