How to use substitution to solve this integral? ∫_0^1dx/(sqrtx+5root(3)x)

1 Answer
Dec 2, 2017

int_0^1 (6u^5*du)/[sqrt(u^6)+5(u^6)^(1/3)]=137-750Ln(6/5)

Explanation:

After using u=x^(1/6), x=u^6 and dx=6u^5*du transforms, this integral became,

int_0^1 (6u^5*du)/[sqrt(u^6)+5(u^6)^(1/3)]

=int_0^1 (6u^5*du)/(u^3+5u^2)

=int_0^1 (6u^3*du)/(u+5)

=int_0^1 ((6u^3+30u^2-30u^2-150u+150u+750-750)*du)/(u+5)

=int_0^1 ([(6u^2-30u+150)*(u+5)-750]*du)/(u+5)

=int_0^1 (6u^2-30u+150)*du-int_0^1 (750*du)/(u+5)

=[2u^3-15u^2+150u]_0^1-[750Ln(u+5)]_0^1

=137-750Ln(6/5)