How do you evaluate the integral #int dx/(root3(x)+1)#?

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Answer 1 · 2018-04-23T00:03:07

The integral is equal to #3/2(x^(2/3)-2x^(1/3)+2ln|x^(1/3)+1|)+C#.

#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!