Integrating x^3/(1+x^2) from 0 to 1.48766439

  • Thread starter Thread starter Swerting
  • Start date Start date
  • Tags Tags
    Bounded Integral
Swerting
Messages
35
Reaction score
0

Homework Statement


I have to take an integral of x^3/(1+x^2) from zero to 1.48766439...(I have the number).


Homework Equations


None really.


The Attempt at a Solution


Well, I tried and tried, and I could not find a single way to separate the top from the bottom. Also, I tried u substitution of both x^3 and 1+x^2 but it never seemed to work out. I'm not sure if this would be good for integration of parts, since I believe that one must be able to be integrated multiple times, such as e^x, so any nudge in the right direction would help. I have the correct answer, I just would like to be able to know how to get to it. Thank you for your time.

-Swerting
 
Last edited:
Physics news on Phys.org
Try writing it as x^2*x/(1+x^2). Now substitute u=1+x^2. Replace the x^2 in the numerator by u-1. Do you see it now?
 
Ah yes! I completely forgot about that! Thank you very much, I do believe I have it now!
 
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