color(white)=int1/(root3x+1) dx
=int1/(x^(1/3)+1) dx
To solve the integral, substitute u=x^(1/3)+1, which means:
du=1/3x^(-2/3)dx
du=1/(3x^(2/3))dx
3x^(2/3)du=dx
But we can solve for x^(2/3) using our original substitution:
u=x^(1/3)+1
u-1=x^(1/3)
(u-1)^2=x^(2/3)
Put this in the other equation:
3(u-1)^2du=dx
Plug this into the integral:
color(white)=int1/u*3(u-1)^2du
=3int(u-1)^2/u du
=3int(u^2-2u+1)/u du
=3int(u-2+1/u) du
=3(intu du-int2 du+int1/udu)
=3(u^2/2-2u+ln|u|)+C
=(3u^2)/2-6u+3ln|u|+C
=(3(x^(1/3)+1)^2)/2-6(x^(1/3)+1)+3ln|x^(1/3)+1|+C
=(3(x^(2/3)+2x^(1/3)+1))/2-6(x^(1/3)+1)+3ln|x^(1/3)+1|+C
=(3x^(2/3)+6x^(1/3)+3)/2-6x^(1/3)+6+3ln|x^(1/3)+1|+C
=(3x^(2/3)+6x^(1/3))/2-6x^(1/3)+3ln|x^(1/3)+1|+C+3/2+6
=(3x^(2/3)+6x^(1/3)-12x^(1/3)+6ln|x^(1/3)+1|)/2+C
=(3x^(2/3)-6x^(1/3)+6ln|x^(1/3)+1|)/2+C
=(3(x^(2/3)-2x^(1/3)+2ln|x^(1/3)+1|))/2+C
=3/2(x^(2/3)-2x^(1/3)+2ln|x^(1/3)+1|)+C
That's the integral. Hope this helped!