Can i get some help with this integral

  • Thread starter Thread starter johnnyboy2005
  • Start date Start date
  • Tags Tags
    Integral
johnnyboy2005
Messages
29
Reaction score
0
from negative infinity to infinity (1/ 4x^2 + 4x +5) dx


is there a way to simplify with partial fracitons or should i do something else? thanks for the help.
 
Physics news on Phys.org
Do you mean:

<br /> \int_{ - \infty }^\infty {\left[ {\frac{1}<br /> {{4x^2 }} + 4x + 5} \right]} \,dx<br />

or

<br /> \int_{ - \infty }^\infty {\left[ {\frac{1}<br /> {{4x^2 + 4x + 5}}} \right]\,dx} <br />

If you meant the first, you can perform integration on each term independently, and add the resulting terms.

If you meant the second, you should look at an integral table, and find those dealing with arctangents.

http://functions.wolfram.com/ElementaryFunctions/ArcTan/07/01/01/

- Warren
 
Complete the square, if the second option is the right one.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top