# Apostol (6.25.40): Find $\int {x^{-2}}\sqrt{2 - x - x^2} dx$

I've been struggling with this exercise from Apostol for some time (Section 6.25, Question 40). The integral is $$\int\frac{\sqrt{2-x-x^2}}{x^2}\, dx$$ with a Hint of "multiply numerator and denominator by $\sqrt{2-x-x^2}$".

NB: the answer supplied in the book is $$-\frac{\sqrt{2-x-x^2}}{x} +\frac{\sqrt{2}}{4}\log(\frac{\sqrt{2-x-x^2}}{x} - \frac{\sqrt{2}}{4}) - \arcsin\frac{2x+1}{3} + C$$

I performed the hint and I obtained the following: $$\int\frac{\sqrt{2-x-x^2}}{x^2}\, dx = \int\frac{\sqrt{2-x-x^2}}{x^2}\cdot \frac{\sqrt{2-x-x^2}}{\sqrt{2-x-x^2}}\cdot dx$$ $$= \int\frac{2-x-x^2}{x^2\sqrt{2-x-x^2}}\cdot dx$$ $$= \int\frac{2}{x^2\sqrt{2-x-x^2}}\cdot dx - \int\frac{x}{x^2\sqrt{2-x-x^2}}\cdot dx- \int\frac{x^2}{x^2\sqrt{2-x-x^2}}\cdot dx$$ $$= \int\frac{2}{x^2\sqrt{2-x-x^2}}\cdot dx - \int\frac{dx}{x\sqrt{2-x-x^2}} - \int\frac{dx}{\sqrt{2-x-x^2}}$$ Now I can easily show that $$\int\frac{dx}{\sqrt{2-x-x^2}} = \arcsin{\frac{2x+1}{3}}$$ I have not tackled the integral $\int\frac{2}{x^2\sqrt{2-x-x^2}}\cdot dx$ though I suspect a substitution of $x+0.5 = 1.5 \sin{t}$ would help since I already know that $2-x-x^2 = (\frac{3}{2})^2 - (x+\frac{1}{2})^2$. The integral I am struggling with (at the moment) is $\int\frac{dx}{x\sqrt{2-x-x^2}}$.

So far I have tried the following:

1. An initial substitution of $x+\frac{1}{2} = \frac{3}{2}\sin(t)$ which resulted in the following: $$\int\frac{dx}{x\sqrt{2-x-x^2}} = -2\int\frac{dt}{1-3\sin(t)}$$
2. A second substitution of $u = \tan\frac{t}{2}$ gave the following: $$-2\int\frac{dt}{1-3\sin(t)} = -4\int\frac{du}{u^2-6u+1}$$
3. Using integration by partial fractions, I can show that $$-4\int\frac{du}{u^2-6u+1} = \frac{-1}{\sqrt{2}}\log|\frac{u-3-2\sqrt{2}}{u-3+2\sqrt{2}}|$$
4. This would mean that $$-2\int\frac{dt}{1-3\sin(t)} = \frac{-1}{\sqrt{2}}\log|\frac{\tan\frac{t}{2}-3-2\sqrt{2}}{\tan\frac{t}{2}-3+2\sqrt{2}}|$$
5. Since $x+\frac{1}{2} = \frac{3}{2}\sin(t)$ implies $\cos(t) = \frac{2}{3}\sqrt{2-x-x^2}$ and $\tan\frac{t}{2} = \frac{1-\cos(t)}{\sin(t)}$ we start to get something that is clearly not looking like anything close to the answer

So my questions that I am looking for answers are:

1. Is this the right approach for this integral or am I overlooking something more obvious?; and
2. Will a similar approach be required for the integral $\int\frac{2}{x^2\sqrt{2-x-x^2}}\cdot dx$ ?
• eww.. this is quite ugly – Sandeep Silwal Aug 12 '14 at 3:51
• I forgot Apostol was such a mean guy. – Pedro Tamaroff Aug 12 '14 at 3:52
• Agreed @SandeepSilwal that's why I can't help wondering if there is some more obvious observation to this. – emjay Aug 12 '14 at 3:55
• It has been my observation with Apostol that he has many exercises that are ugly on the surface but break down nicely with some observations. If that is the case here, I'm clearly not seeing it. – emjay Aug 12 '14 at 4:00

For the integral $\int\dfrac{dx}{x\sqrt{2-x-x^2}}$, we can substitute $x=\dfrac{1}{t}$, so that $dx=-\dfrac{1}{t^2}dt$ and the integral becomes $\int\dfrac{-\frac{1}{t^2}dt}{\frac{1}{t}\sqrt{2-\frac{1}{t}-\frac{1}{t^2}}}=-\int\dfrac{dt}{\sqrt{2t^2-t-1}}$ which can be evaluated easily. I think this should work out.