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

[lfdahl]

Find the antiderivative, $F$, of the function $$f(x) = \frac{\sqrt{2-x-x^2}}{x^2}.$$

 

[lfdahl]

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:

Spoiler

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:

\[F = 2^{-\frac{3}{2}}\ln \left ( \left | \frac{4}{x} - \frac{2^{\frac{3}{2}}\sqrt{-x^2-x+2}}{x} - 1\right | \right )-\frac{\sqrt{-x^2-x+2}}{x}+\arctan\left (\sqrt{ \frac{2+x}{1-x} }\right )+C’\]

 

[Opalg]

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)