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Question is self explanatory. I have an exam and our professor gave us questions. This is the one I couldn't do. Any ideas would be very helpful:

$$\int \frac{\sqrt{-x^2 - x + 2}}{x^2}dx$$

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    $\begingroup$ Ugh, I think this is a bit overcomplicated for an exam. Anyways, it might be a good start to integrate by parts (integrate $1/x^2$). $\endgroup$ – mickep May 22 '15 at 18:55
  • $\begingroup$ $-x^2-x+2=\frac94-\left(x+\frac 12\right)^2$ $\endgroup$ – abiessu May 22 '15 at 18:56
  • $\begingroup$ We learned partial integration, substitution of variables, integration of trigonometric functions and integration of irrational functions. $\endgroup$ – Bora Semiz May 22 '15 at 18:56
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    $\begingroup$ Looking at Wolfram's solution, it looks like you want to start with integration by parts, with the piece being integrated being $1/x^2$. Then (matching up with abiessu's suggestion) take $u=x+1/2$, hence $x=u-1/2$, and work from there. It looks like you use partial fractions at that stage, in the process getting a log and an inverse sine function. That...does sound rather horrifying for an exam, I certainly never had anything this hard on a calculus exam. $\endgroup$ – Ian May 22 '15 at 18:58
  • $\begingroup$ Hint: Integrate by parts, then complete the square in the integrand of the resulting integral, and then make an approrpiate trig substitution. $\endgroup$ – Matematleta May 22 '15 at 19:08
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Hint: set $$\sqrt{-x^2-x+2}=xt+\sqrt{2}$$ we get $$x=\frac{-2t\sqrt{2}-1}{t^2+1}$$ and $$dx=\frac{2(-\sqrt{2}+t+\sqrt{2}t^2)}{(1+t^2)^2}dt$$

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    $\begingroup$ Could you please elaborate ? I tried what you suggest but I arrived to a monster which, to me, looks worse than the original integrand. Thanks. $\endgroup$ – Claude Leibovici May 23 '15 at 5:34
  • $\begingroup$ This reads like the suggestion of a CAS gone mad. (And CAS do not answer query in comments either.) $\endgroup$ – Did May 24 '15 at 17:29
  • $\begingroup$ this is the so called Euler substitution $\endgroup$ – Dr. Sonnhard Graubner May 24 '15 at 19:14

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