Error while integrating reciprocal of irreducible quadratic $\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$

Question

I am trying to derive a formula for indefinite integral of reciprocal of irreducible real quadratic. $$\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$$, $b^2 - 4ac \lt 0$. According to WolframAlpa it should come out as: $$\frac{2\arctan{\frac{2ax+b}{\sqrt{4ac - b^2}}}}{\sqrt{4ac - b^2}} + C$$ Instead I get $$\frac{\arctan{\frac{x + \frac{b}{2a}}{\sqrt{\frac{c}{a} - (\frac{b}{2a})^2}}}}{a(\frac{c}{a} - (\frac{b}{2a})^2)^\frac{3}{2}} + C$$ which, no matter how I try to simplify it, doesn't give the right answer. Here are my steps, what am I doing wrong?

1) Reduce to a monic quadratic and complete the square: $$\int{\frac{1}{ax^2 + bx + c}}{dx} = \frac{1}{a}\int{\frac{1}{x^2 + \frac{b}{a}x + \frac{c}{a}}}{dx} = \frac{1}{a}\int{\frac{1}{(x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2}}{dx}$$
3) Substitute $x = u - \frac{b}{2a}$, $u = x + \frac{b}{2a}$, $k = \frac{c}{a} - (\frac{b}{2a})^2$: $$\frac{1}{a}\int{\frac{1}{(x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2}}{dx} = \frac{1}{a}\int{\frac{1}{u^2 + k}}{du}$$
4) Substitute $u = v\sqrt{k}$, $v = \frac{u}{\sqrt{k}}$, $du = \frac{1}{\sqrt{k}}\hphantom ddv$: $$\frac{1}{a}\int{\frac{1}{u^2 + k}}{du} = \frac{1}{a}\int{\frac{\frac{1}{\sqrt{k}}}{kv^2 + k}}{dv} = \frac{1}{a}\int{\frac{1}{k\sqrt{k}(v^2 + 1)}}{dv} = \frac{1}{ak^\frac{3}{2}}\int{\frac{1}{v^2 + 1}}{dv}$$ 5) Use common integral $$\frac{1}{ak^\frac{3}{2}}\int{\frac{1}{v^2 + 1}}{dv} = \frac{\arctan{v}}{ak^\frac{3}{2}} + C$$ 6) Substitute back $$\frac{\arctan{v}}{ak^\frac{3}{2}} + C = \frac{\arctan{\frac{u}{\sqrt{k}}}}{ak^\frac{3}{2}} + C = \frac{\arctan{\frac{x + \frac{b}{2a}}{\sqrt{\frac{c}{a} - (\frac{b}{2a})^2}}}}{a(\frac{c}{a} - (\frac{b}{2a})^2)^\frac{3}{2}} + C$$

Answer 1

Everything is fine till you write $du=\frac{1}{\sqrt{k}}dv$. It is instead, $du=\sqrt{k}dv$. Aside from that, your solution method is great.

Also, for the argument to the arctan function, note that $\frac{x+\frac{b}{2a}}{\sqrt{\frac{c}{a}-\left(\frac{b}{2a}\right)^2}}=\frac{2a\left(x+\frac{b}{2a}\right)}{2a\sqrt{\frac{c}{a}-\left(\frac{b}{2a}\right)^2}}$, and simplifying this will allow the "inside" of the tangent function to have the appropriate form.

Answer 2

The error is in step 4: if $u=\sqrt kv$, then $\mathrm du=\sqrt k\mathrm dv$, and not $\frac1{\sqrt k}\,\mathrm dv$.

Answer 3

In my opinion you can save many hassles with a slightly different method. Set $\sqrt{4ac-b^2}=D$ and do \begin{align} \int\frac{1}{ax^2+bx+c}\,dx &=\int\frac{4a}{4a^2x^2+4abx+4ac}\,dx \\[6px] &=\int\frac{4a}{(2ax+b)^2+(4ac-b^2)}\,dx \\[6px] &=\int\frac{4a}{D^2t^2+D^2}\frac{D}{2a}\,dt && 2ax+b=Dt,\quad dx=\frac{D}{2a}\,dt \\[6px] &=\frac{2}{D}\int\frac{1}{t^2+1}\,dt \\[6px] &=\frac{2}{D}\arctan t+C \\[6px] &=\frac{2}{\sqrt{4ac-b^2}}\arctan\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right) \end{align}

Note that if you set $u=Kv$, then $du=K\,dv$, not $du=\frac{1}{K}\,dv$. For instance, in the substitution above, $$ 2a\,dx=D\,dt $$ so $$ dx=\frac{D}{2a}\,dt $$