int 1/(sqrt(x)-root3(x)) dx Use substitution and partial fractions to find the indefinite integral

int1/(sqrt(x)-root(3)(x))dx
Apply integral substitution:u=x^(1/6)
=>du=1/6x^(1/6-1)dx
du=1/6x^(-5/6)dx
du=1/(6x^(5/6))dx
6x^(5/6)du=dx
6(x^(1/6))^5du=dx
6u^5du=dx
int1/(sqrt(x)-root(3)(x))dx=int(6u^5)/(u^3-u^2)du
=int(6u^5)/(u^2(u-1))du
Take the constant out,
=6intu^3/(u-1)du
Integrand is an inproper rational function as degree of numerator is more than the degree of the denominator,
So let's carry out the division,
u^3/(u-1)=u^2+u+1+1/(u-1)
=6int(u^2+u+1+1/(u-1))du
Apply the sum rule,
=6(intu^2du+intudu+int1du+int1/(u-1)du)
Apply the power rule and the common integer:int1/xdx=ln|x|
=6(u^3/3+u^2/2+u+ln|u-1|)
Substitute back u=x^(1/6)
and add a constant C to the solution,
=6(1/3(x^(1/6))^3+1/2(x^(1/6))^2+x^(1/6)+ln|x^(1/6)-1|)+C
=2x^(1/2)+3x^(1/3)+6x^(1/6)+6ln|x^(1/6)-1|+C
=2sqrt(x)+3root(3)(x)+6root(6)(x)+6ln|root(6)(x)-1|+C