color(white)=int1/(root3x+1)=∫13√x+1 dxdx
=int1/(x^(1/3)+1)=∫1x13+1 dxdx
To solve the integral, substitute u=x^(1/3)+1u=x13+1, which means:
du=1/3x^(-2/3)dxdu=13x−23dx
du=1/(3x^(2/3))dxdu=13x23dx
3x^(2/3)du=dx3x23du=dx
But we can solve for x^(2/3)x23 using our original substitution:
u=x^(1/3)+1u=x13+1
u-1=x^(1/3)u−1=x13
(u-1)^2=x^(2/3)(u−1)2=x23
Put this in the other equation:
3(u-1)^2du=dx3(u−1)2du=dx
Plug this into the integral:
color(white)=int1/u*3(u-1)^2du=∫1u⋅3(u−1)2du
=3int(u-1)^2/u=3∫(u−1)2u dudu
=3int(u^2-2u+1)/u=3∫u2−2u+1u dudu
=3int(u-2+1/u)=3∫(u−2+1u) dudu
=3(intu=3(∫u du-int2du−∫2 du+int1/udu)du+∫1udu)
=3(u^2/2-2u+ln|u|)+C=3(u22−2u+ln|u|)+C
=(3u^2)/2-6u+3ln|u|+C=3u22−6u+3ln|u|+C
=(3(x^(1/3)+1)^2)/2-6(x^(1/3)+1)+3ln|x^(1/3)+1|+C=3(x13+1)22−6(x13+1)+3ln∣∣x13+1∣∣+C
=(3(x^(2/3)+2x^(1/3)+1))/2-6(x^(1/3)+1)+3ln|x^(1/3)+1|+C=3(x23+2x13+1)2−6(x13+1)+3ln∣∣x13+1∣∣+C
=(3x^(2/3)+6x^(1/3)+3)/2-6x^(1/3)+6+3ln|x^(1/3)+1|+C=3x23+6x13+32−6x13+6+3ln∣∣x13+1∣∣+C
=(3x^(2/3)+6x^(1/3))/2-6x^(1/3)+3ln|x^(1/3)+1|+C+3/2+6=3x23+6x132−6x13+3ln∣∣x13+1∣∣+C+32+6
=(3x^(2/3)+6x^(1/3)-12x^(1/3)+6ln|x^(1/3)+1|)/2+C=3x23+6x13−12x13+6ln∣∣x13+1∣∣2+C
=(3x^(2/3)-6x^(1/3)+6ln|x^(1/3)+1|)/2+C=3x23−6x13+6ln∣∣x13+1∣∣2+C
=(3(x^(2/3)-2x^(1/3)+2ln|x^(1/3)+1|))/2+C=3(x23−2x13+2ln∣∣x13+1∣∣)2+C
=3/2(x^(2/3)-2x^(1/3)+2ln|x^(1/3)+1|)+C=32(x23−2x13+2ln∣∣x13+1∣∣)+C
That's the integral. Hope this helped!
