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