Question #eab38 Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Alberto P. 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)# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 2529 views around the world You can reuse this answer Creative Commons License