[int from 0 to 1] x^4(1-x)^4/(1+x^2)

[dts5044]

The problem is simple to integrate [integral from 0 to 1] of
x[sup:24l4dwyr]4[/sup:24l4dwyr](1-x)[sup:24l4dwyr]4[/sup:24l4dwyr] * dx
---------------------
1 + x[sup:24l4dwyr]2[/sup:24l4dwyr]


I've tried multiplying out and didn't see any solution. I tried a trig substitution and didn't seem to get anywhere either. Can anyone help?

 

[stapel]

I've tried multiplying out and didn't see any solution.

Try multiplying out the numerator, and then doing the long division. I think you should end up with the integrand being:

. . . . .\(\displaystyle x^4\, -\, 4x^3\, +\, 5x^2\, -\, 4\, +\, \frac{4}{1\, +\, x^2}\)

...which should be a bit easier to deal with. :wink:

Eliz.

 

[Dr. Flim-Flam]

Integral [x^4(1-x)^4]/(1+x^2),x,0,1 = integral [-4/(x^2+1) + x^6 -4x^5 + 5x^4 - 4x^2 + 4],x,0,1

= -4arctan(x) +x^7/7 - 2x^6/3 + x^5 -4x^3/3 + 4x],0,1 = [22-7Pi]/7

Note, since the binomial is only of degree 4, might as well expand and then collect terms.

 

[dts5044]

ah, I can't believe I missed that. Thank you!