What is the integral $\displaystyle\int_{-\infty}^{+\infty}\frac{xdx}{(x^2 + bx + c)^{3/2}}$?

Question

Where $b,c$ are constants.This seems to be difficult to find analytically.

Answer 1

We have $$I =\int_{-\infty}^{\infty} \frac {x}{(x^2+bx+c)^{\frac {3}{2}}} dx =\frac {1}{2} \int_{-\infty}^{\infty} \frac {2x+b}{(x^2+bx+c)^{\frac {3}{2}}} dx - \frac {b}{2} \int_{-\infty}^{\infty} \frac {1}{(x^2+bx+c)^{\frac {3}{2}}} dx =\frac {1}{2} I_1- \frac {b}{2} I_2$$

To calculate $I_2$, complete the square in the denominator and substitute $u=2x+b $. Computing $I_1$is relatively easy. The answer is $$\boxed {-\frac {4b}{4c-b^2}} $$ assuming $4c-b^2>0$. Hope it helps.

Answer 2

HINT:

$$x^2+bx+c=\dfrac{(x+2b)^2+4c-4b^2}4$$

Using Trigonometric substitutions,

if $4c-4b^2>0,$ choose $x+2b=\sqrt{4c-4b^2}\tan t$

if $4c-4b^2<0,$ choose $x+2b=\sqrt{4b^2-4c}\sec t$

if $4c-4b^2=0$ can you try?