Question

qual integral de (9t^2+1/raiz quadrada t^3)dt

Answer 1

Resposta:

$\displaystyle\int \frac{9t^2+1}{\sqrt{t^3} } dt= \displaystyle\int(\frac{9t^2}{\sqrt{t^3} } +\frac{1}{\sqrt{t^3} })dt =\displaystyle\int9t^2t^{-\frac{3}{2} } dt+\displaystyle\int t^{-\frac{3}{2} }dt=9\displaystyle\int t^{\frac{1}{2} } dt+\displaystyle\int t^{-\frac{3}{2} } dt}$

$9*\frac{t{} ^{\frac{3}{2} } }{\frac{3}{2} } +\frac{t^{-\frac{1}{2} } }{-\frac{1}{2} } +c=9*\frac{2}{3} t^{\frac{3}{2} } -2t^{- \frac{1}{2} } +c=6\sqrt{t^3} -\frac{2}{\sqrt{t} }+c$