$\int{\frac{1}{(\alpha x^2+\beta x+\gamma)^2}\;dx}$, where $\beta^2-4\alpha\gamma < 0$

Question

I'm trying to solve this partial fraction integral $$\int{\frac{1}{(\alpha x^2+\beta x+\gamma)^2}\;dx}$$ where $a\neq 0$ and $\beta^2-4\alpha\gamma< 0$.

This is an irreducible polynomial with a raised power.

So far I have tried to apply it into the form $$\frac{1}{(\alpha x^2+\beta x+\gamma)^2}=\frac{Ax+B}{(\alpha x^2+\beta x+\gamma)}+\frac{Cx+D}{(\alpha x^2+\beta x+\gamma)^2}$$ $$1=(Ax+B)(\alpha x^2+\beta x+\gamma)+(Cx+D)$$ I have tried to proceed with $A,B,C=0$ and $D=1$, but it brought me back to $\frac{1}{(\alpha x^2+\beta x+\gamma)^2}$.

Also I think I can take the $\alpha$ out as: $$\frac{Ax+B}{\alpha x^2+\beta x+\gamma}=\frac{1}{a}\frac{Ax+B}{x^2+\beta \frac{x}{a}+\frac{\gamma}{a}}$$ but don't know how to proceed with a raised power 2.

Answer 1

Hint.

• Assume $\beta^2-4\alpha\gamma< 0$. Then one may write $$ \begin{align} \int{\frac{1}{(\alpha x^2+\beta x+\gamma)^2}\;dx}&=\frac1{\alpha^2}\int{\frac{1}{\left[\left(x+\frac{\beta}{2\alpha}\right)^2+\left(\frac{\sqrt{4\alpha\gamma-\beta^2}}{2\alpha}\right)^2\right]^2}\;dx} \\\\&=\frac1{\alpha^2}\int{\frac{du}{\left[u^2+c^2\right]^2}}, \quad \left(u=x+\frac{\beta}{2\alpha},\,c=\frac{\sqrt{4\alpha\gamma-\beta^2}}{2\alpha}\right) \\\\&=\frac1{\alpha^2c}\int{\cos^2\theta}\:d\theta, \quad u=c\,\tan \theta. \end{align} $$ Hope you can finish it.
• The case $\beta^2-4\alpha\gamma= 0$ gives $$ \int{\frac{1}{(\alpha x^2+\beta x+\gamma)^2}\;dx}=\frac1{\alpha^2}\int{\frac{dx}{\left(x+\frac{\beta}{2\alpha}\right)^4}} $$ which is manageable.