0
$\begingroup$

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

  • 0
    Did you try anything. It does not look impossible. The only problem I see is that the polynomial $x^2+bx+c$ must not be zero, since then the integral is divergent. – 2017-01-22
  • 0
    Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, [this link](http://meta.math.stackexchange.com/a/9960) might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – 2017-01-26

2 Answers 2

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.

  • 0
    Wont I2 be 0? The integral has the same value 1/(c- (b^2)/4) for both the limits. – 2017-01-22
1

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?