Question #eab38

1 Answer
Nov 11, 2016

#6(x^(7/6)/7-x^(5/6)/5+x^(1/2)/3-x^(1/6)+arctanx^(1/6))+c#

Explanation:

Use #z=x^(1/6)\ \ => x=z^6, sqrt(x)=z^3, root(3)(x)=z^2, dx=6z^5dz#
#intsqrtx/(1+root(3)x)dx=6intz^8/(1+z^2)dz#

Then

#z^8/(1+z^2)=z^6-z^4+z^2-1+1/(a+z^2)#

so

#intsqrtx/(1+root(3)x)dx=6(z^7/7-z^5/5+z^3/3-z+arctan z)+c#

then make substitution #z=x^(1/6)#