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Is there someone who can show me how to evaluate this integral:

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

I have tried many changes of variables but I haven't succeeded yet. Thank you for any help.

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  • $\begingroup$ it could be by integration by part. $\endgroup$ Jul 12, 2015 at 16:23

2 Answers 2

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HINT:

$$-x^2-x+2=\dfrac{3^2-(2x+1)^2}4$$

Set $2x+1=3\sin\theta$

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$\begin{gathered} \int {\frac{{\sqrt { - {x^2} - x + 2} }}{{{x^2}}}} \,dx \hfill \\ = - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \int {\frac{{\left( { - 2x - 1} \right)}}{{2\sqrt { - {x^2} - x + 2} }}} \frac{{ - 1}}{x}\,dx\quad \quad \left( {{\text{Integrating by parts}}} \right) \hfill \\ = - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \int {\frac{{2x + 1}}{{2x\sqrt { - {x^2} - x + 2} }}} \,dx \hfill \\ = - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \left[ {\int {\frac{{dx}}{{\sqrt {\left( {2 + x} \right)\left( {1 - x} \right)} }} + \int {\frac{{dx}}{{2x\sqrt { - {x^2} - x + 2} }}} } } \right] \hfill \\ = - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \left[ {2{{\sin }^{ - 1}}\sqrt {\frac{{x + 2}}{3}} - \frac{1}{{2\sqrt 2 }}\log \left| {\frac{1}{{x\sqrt 2 }} - \frac{1}{{4\sqrt 2 }} + \sqrt {\frac{1}{{2{x^2}}} - \frac{1}{{4x}} - \frac{1}{4}} } \right|} \right]+c \hfill \\ \end{gathered} $

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