How to integrate a partial fraction problem of the form $\frac{(ax+b)}{(ax^2 + bx + c)^n}$?

Question

Please help me integrating a partial fraction problem of the form $$\frac{(ax+b)}{(ax^2 + bx + c)^n}$$

My professor gave this explanation which loses me at $Step-2$.

Can anyone explain this to me in plain English? Thanks in Advance!

Answer 1

So this given solution has a lot of things going on. Hopefully the following will give some clarity.

We want to integrate $\frac{Ax + B}{(ax^2 + bx + c)^n}$ with $b^2 - 4ac < 0$. We first simplify our calculations by factoring out $a^n$ to get $\frac{A'x + B'}{(x^2 + \frac{b}{a}x + \frac{c}{a})^n}$ where $A' = A / a^n$ and $B' = B/a^n$. Now we can complete the square as follows: $$x^2 + (b/a)x + (c/a) = \left(x + \frac{b}{2a}\right)^2 + \frac{c}{a} - \frac{b^2}{4a^2}.$$ Since $b^2 - 4ac < 0$ we have $$\frac{c}{a} - \frac{b^2}{4a^2} = \frac{1}{4a^2}(4ac - b^2) > 0.$$ Now let $r^2 = \frac{c}{a} - \frac{b^2}{4a^2}$ and $rt = x + b/2a$ so $dx = r dt$. Now our integral becomes $$\int \frac{A'x + B'}{(x^2 + \frac{b}{a}x + \frac{c}{a})^n}dx = \int\frac{A'(rt - b/2a) + B'}{((rt)^2 + r^2)^n}rdt = \int \frac{(A'r^2/r^{2n})t + (B'r - rb/2a)/r^{2n}}{(t^2 +1)^n}dt = \int\frac{Ct + D}{(t^2 + 1)}dt$$ where $C = A'r^2/r^{2n}$ and $D = (B'r - rb/2a)/r^{2n}$.

Let me know if you need to clarify anything further or explain the rest of the solution.