Another messy integral: $I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$ I found the following question in a practice book of integration:-
$Q.$ Evaluate $$I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$$ For this I substituted $t^2=\frac {2-x-x^2}{x^2}\implies x^2=\frac{2-x}{1+t^2}\implies 2t\ dt=\left(-\frac4{x^3}+\frac 1{x^2}\right)\ dx$. Therefore $$\begin{align}I&=\int\frac {\sqrt{2-x-x^2}}{x^2}\ dx\\&=\int \left(\frac tx\right)\left(\frac{2t\ dt}{-\frac4{x^3}+\frac 1{x^2}}\right)\\&=\int \frac{2t^2\ dt}{\frac{x-4}{x^2}}\\&=\int \frac{2t^2(1+t^2)\ dt}{{x-4}\over{2-x}}\\&=\int \frac{2t^2(1+t^2)(5+4t^2-\sqrt{8t^2+9})\ dt}{\sqrt{8t^2+9}-(8t^2+9)}\end{align}$$ Now I substituted $8t^2+9=z^2 \implies t^2=\frac {z^2-9}8 \implies 2t\ dt=z/4\ dz$. So, after some simplification, you get $$\begin{align}I&=-\frac1{512}\int (z^2-9)(z+1)(z-1)^2\ dz\end{align}$$ I didn't have the patience to solve this integration after all these substitutions knowing that it can be done (I think I have made a mistake somewhere but I can't find it. There has to be an $ln(...)$ term, I believe). Is there an easier way to do this integral, something that would also strike the mind quickly? I have already tried the Euler substitutions but that is also messy.

Hint: $\sqrt{2-x-x^2}=\sqrt{\frac94-(x+\frac12)^2}$ 
Substitute $x+\frac12=t; dx=dt$ 
$$\int \frac{\sqrt{\frac94-t^2}}{(t-\frac12)^2}\ dt$$
Put $\frac32\sin\theta=t;\frac32\cos\theta d\theta=dt$
$$\int \frac{\frac32\cos\theta.\frac32\cos\theta d\theta}{(\frac32\sin\theta-\frac12)^2}$$
$$9\int \frac{\cos^2\theta d\theta}{(9\sin^2\theta+1-6\sin \theta)}$$
Now do $u=\tan\frac{\theta}2;du=\frac12\sec^2{\frac{\theta}2}d\theta$
$$\boxed{\sin\theta=\frac{2u}{1+u^2};\cos\theta=\frac{1-u^2}{1+u^2};d\theta=\frac{2du}{u^2+1}}$$
I guess you can take it from here?
