How do you integrate this? int_0^1(x^4(1-x)^4)/(1+x^2)dx

1 Answer
Mar 17, 2018

int_0^1(x^4(1-x)^4)/(1+x^2)dx=22/7-pi

Explanation:

Let

I=int_0^1(x^4(1-x)^4)/(1+x^2)dx

Expand the numerator:

I=int_0^1(x^8-4x^7+6x^6-4x^5+x^4)/(x^2+1)dx

Apply long division:

I=int_0^1(x^6-4x^5+5x^4-4x^2-4/(x^2+1)+4)dx

Integrate directly:

I=[1/7x^7-2/3x^6+x^5-4/3x^3-4tan^(-1)x+4x]_0^1

Hence

I=22/7-pi