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I found an integral from the MIT Integration Bee. The following problem was supposed to be solved within a 2 minute time limit.

$$\displaystyle\int \dfrac{\sqrt{x^3-1}}{x}\,dx$$

I understand that this problem may be solved using hyperbolic trigonometric functions, but is there a quicker solution (since you are only given 2 minutes to solve the integral)?

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up vote 7 down vote accepted

Let $u^2 = x^3-1$, $2u \, du = 3x^2 \, dx$, so we have $$\int \frac{\sqrt{x^3 - 1}}{x} \, dx = \frac{1}{3} \int 3x^2 \frac{\sqrt{x^3-1}}{x^3} \, dx = \frac{2}{3} \int \frac{u^2}{u^2+1} \, du,$$ and the rest is easy.

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Ohh.. the pressure, to do it in 2 mins. You made it though. – Rene Schipperus Jun 9 '14 at 19:52
This was an excellent solution. Thank you. – A is for Ambition Jun 9 '14 at 19:58
@Lucian No. My computation is correct. The extra $u$ comes from the substitution. – heropup Jun 9 '14 at 21:46

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