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I have been trying to solve this one but I have no clue, I put it into wolfram and the result is absurd, considering this one is taken from MIT bee qualifier 2012, this is the integral: $$\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \, dx$$

Michael Hardy
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Juan Pablo Arcila
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  • Why such a negative reaction to this question ? – Rene Schipperus Dec 03 '17 at 05:04
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    What have you attempted? Include it in the post so it doesn't get closed – Dylan Dec 03 '17 at 05:05
  • @Dylan Why is it necessary to include attempts to avoid closure ? – Rene Schipperus Dec 03 '17 at 05:06
  • I agree with Rene, I think given that this is a contest math question a downvote and a close vote due to not showing an attempt is less appropriate than a homework question. Nevertheless it is more helpful to note that $I = 2\tan^{-1}\sqrt{x + \frac{1}{x} + 1}$ rather than the Wolfram Alpha output. – JessicaK Dec 03 '17 at 05:07
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    I don't know, I'm just following from past questions that have been closed due to lack of attempt. Personally I don't even mind homework questions. – Dylan Dec 03 '17 at 05:13
  • @Dylan So why enforce something you dont agree with ? – Rene Schipperus Dec 03 '17 at 05:15
  • Because I want to help OPs avoid having the questions closed, like what's happened to me in the past. I have no control over how the rules are created. Can't beat them, join them. – Dylan Dec 03 '17 at 05:22

2 Answers2

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$$ \begin{align} \int\frac{x-1}{x+1}\frac{\mathrm{d}x}{\sqrt{x^3+x^2+x}} &=\int\frac{x^{-1/2}-x^{-3/2}}{x^{1/2}+x^{-1/2}}\frac{\mathrm{d}x}{\sqrt{\left(x^{1/2}+x^{-1/2}\right)^2-1}}\\ &=2\int\frac1{x^{1/2}+x^{-1/2}}\frac{\mathrm{d}\!\left(x^{1/2}+x^{-1/2}\right)}{\sqrt{\left(x^{1/2}+x^{-1/2}\right)^2-1}}\\ &=2\int\frac1u\frac{\mathrm{d}u}{\sqrt{u^2-1}}\\[9pt] &=2\sec^{-1}(u)+C\\[15pt] &=2\sec^{-1}\left(x^{1/2}+x^{-1/2}\right)+C \end{align} $$

robjohn
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You can rewrite the integrand as

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

Then make the substitution $u^2 = x + \dfrac{1}{x} + 1$ to get

$$ \int \frac{2}{u^2+1} du = 2\arctan u + C $$

Dylan
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