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

1 Answer
Apr 23, 2018

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

Explanation:

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!