Calculus: Early Transcendentals, Chapter 7, Chapter Review, Section Review, Problem 28

int (root(3)(x)+1)/(root(3)(x)-1) dx
The above indefinite integral can be rewritten as:
= int ((root(3)(x)+1-(root(3)(x)-1))/(root(3)(x)-1)+1) dx
Applying the sum rule:
= int 2/(root(3)(x)-1) dx + int 1 dx
Let u=root(3)(x), du=1/(3x^(2/3)) dx, dx=3x^(2/3) du
= 2 int (3x^(2/3))/(u-1) du + int dx
Again, u=root(3)(x) => x= u^3
= 2 int (3u^2)/(u-1) du + int dx
= 2.3 int (u+1/(u-1)+1) du + int dx
Again applying the sum rule:
=6(u^2/2+ln(u-1)+u)+x+C
Substituting back u=root(3)(x) ,
=6(x^(2/3)/2+ln(root(3)(x)-1)+root(3)(x))+x+C
Therefore, the answer is 6(x^(2/3)/2+ln(root(3)(x)-1)+root(3)(x))+x+C