Difficult Integration - Apostol Section 6.25 #40

[process91]

Homework Statement

[itex]\int\frac{\sqrt{2-x-x^2}}{x^2}dx[/itex]

Hint: multiply the numerator and denominator by [itex]\sqrt{2-x-x^2}[/itex]

Homework Equations

This is in the Integration using Partial Fractions section, but the last few have not been using Partial Fractions.

The Attempt at a Solution

Well, initially I thought that I would just complete the square on the top and then use a substitution such as [itex]x+\frac{1}{2} = \frac{3}{2}\sin u[/itex], but that became pretty complex. Then I took the author's suggestion and multiplied the numerator and denominator by [itex]\sqrt{2-x-x^2}[/itex]. I split the resulting integral into three, two of which were easy and the first which is still very difficult:

[itex]\int \frac{2}{x^2 \sqrt{2-x-x^2}} dx[/itex]

The substitution I mentioned earlier still looks most promising, but leads to this:

[itex]\frac{9}{4} \int \frac{\cos^2 u}{(\frac{3}{2}\sin u + \frac{1}{2})^2} du[/itex]

This looks like it needs something like [itex]z=\tan \frac{u}{2}[/itex], but that also becomes extremely tortuous. It leads to a degree 6 polynomial on the bottom, and a degree 4 polynomial on the top which can then be solved with partial fractions, but the resulting equations are quite cumbersome.

Any other suggestions/hints?

 

[ehild]

Try integration by parts:

[itex]9\int \frac{\cos^2 u}{(3\sin u + 1)^2} du=\int (\frac{3\cos u}{(3\sin u+1)^2}) (3 \cos u) du[/itex]

ehild

 

[process91]

That does look much better, thanks!