#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!
