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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 $\int\frac{2}{x^2\sqrt{2-x-x^2}}\cdot dx$ ?
davidlowryduda
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emjay
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1 Answers1

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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.

Debashish
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