MHB Find the antiderivative of V(2−x−x^2)/x^2

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The discussion focuses on finding the antiderivative of the function f(x) = √(2 - x - x²)/x². Participants express that the solution is complex and lacks elegance, making it less appealing compared to other mathematical challenges. One contributor reflects on their initial interest in the problem but ultimately feels relieved not to have pursued it further. The conversation highlights the challenge's intricate nature and the disappointment in its lack of mathematical beauty. Overall, the thread emphasizes the difficulty and unappealing aspects of this particular antiderivative problem.
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Find the antiderivative, $F$, of the function $$f(x) = \frac{\sqrt{2-x-x^2}}{x^2}.$$
 
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The answer to this challenge is neither short nor elegant. In addition the solution path is quite winded.
I´m sorry for having posted the challenge, which seemingly doesn´t contain the mathematical beauty, that
can be found in many other challenges (Sadface)Suggested solution:

From $2-x-x^{2} = (2+x)(1-x)$, it seems a good idea to use the substitution $2+x=y^{2}(1-x).$

Hence, we get:

\[ x=\frac{y^{2}-2}{y^{2}+1}\] and \[dx=\frac{6y}{(y^{2}+1)^{2}}\ dy.\]

and \[\sqrt{2-x-x^{2}}= (1-x)y= \frac{3y}{y^{2}+1}.\]

Thus, our integral/antiderivative can be rewritten in terms of $y$ as:

\[F=\int \frac{3y}{y^{2}+1} \cdot \left (\frac{y^{2}+1}{y^{2}-2} \right )^{2} \cdot \frac{6y}{(y^{2}+1)^{2}}\ dy\]

\[F=\int \frac{18y^{2}}{(y^{2}+1)(y^{2}-2)^{2}}\ dy\]

\[F=\int \left[\frac{2y^{2}+8}{(y^{2}-2)^{2}}-\frac{2}{y^{2}+1}\right]\ dy\]

\[F=\int \frac{2y^{2}+8}{(y^{2}-2)^{2}}\ dy-2\arctan y\]

Using the identity: \[\left(\frac{y}{y^{2}-2}\right)'=-\frac{y^{2}+2}{(y^{2}-2)^{2}}.\]

yields

\[F=\int \frac{3y^{2}+6-(y^{2}-2)}{(y^{2}-2)^{2}}\ dy-2\arctan y\]

\[F=3\int \frac{y^{2}+2}{(y^{2}-2)^{2}}\ dy-\int \frac{1}{y^{2}-2}\ dy-2\arctan y\]

\[F=-\frac{3y}{y^{2}-2}-\int \frac{1}{(y+\sqrt{2})(y-\sqrt{2})}\ dy-2\arctan y\]

\[F=-\frac{3y}{y^{2}-2}-2^{-\frac{3}{2}}\int \left (\frac{1}{y-\sqrt{2}}-\frac{1}{y+\sqrt{2}}\right)\ dy-2\arctan y\]

\[F=-\frac{3y}{y^{2}-2}-2^{-\frac{3}{2}}\ln \left ( \frac{y-\sqrt{2}}{y+\sqrt{2}} \right )-2\arctan y + C\]

Backsubstitution with $y=\sqrt{\frac{2+x}{1-x}}$, yields:

\[F = 2^{-\frac{3}{2}}\ln \left ( \left | \frac{4}{x} - \frac{2^{\frac{3}{2}}\sqrt{-x^2-x+2}}{x} - 1\right | \right )-\frac{\sqrt{-x^2-x+2}}{x}+\arctan\left (\sqrt{ \frac{2+x}{1-x} }\right )+C’\]
 
I had this problem on a list of things to think about when I have some spare time. Having seen the solution, I'm glad I never got round to it. (Shake)
 
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