How to integrate dx/[1+(x)^1/3]^1/2

let u = [1+(x)^1/3]^1/2
and dx would be = 6u(u^2-1)^2 du

so your integral would look like:
∫ ( [6u (u^2-1)^2] / u ) du

= 6 ∫ u^4 -2u^2 +1 du

and you can do the rest :)

thanks for the BA!~
and do your homework next time :p

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