Asked by Vicky

integral of [sqrt(u) - 2 u^2]/u du

i got stuck
∫u^-1 (sqrt(u) - 2 u^2) du
∫ u^(-1/2) - 2u du
i don't know i'm doing right step

13 years ago

Answered by Vicky

ok i got it. don't need ur help anymore

13 years ago

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