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

Question

1270 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-02T12:20:37

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

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)#