Question

a integral de f (x)=$\sqrt{x} (x+ \frac{1}{x} )$ é:

Answer 1

$Fun\c{c}\~ao:\\\\ f_{(x)}=\sqrt{x}(x+\frac{1}{x})\\\\ f_{(x)}=x\sqrt{x}+\frac{\sqrt{x}}{x}\\\\ f_{(x)}=\sqrt{x^3}+\frac{\sqrt{x}}{\sqrt{x}^2}\\\\ f_{(x)}=\sqrt{x^3}+\sqrt{x}^{-1}\\\\ f_{(x)}=x^{\frac{3}{2}}+x^{-\frac{1}{2}}\\\\\\ Integrando:\\\\ F_{(x)}=\displaystyle \int \left(x^{\frac{3}{2}}+x^{-\frac{1}{2}} \right).dx\\\\\\ F_{(x)}=\dfrac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\dfrac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + constante$


$F_{(x)}=\dfrac{x^{\frac{5}{2}}}{\frac{5}{2}}+\dfrac{x^{\frac{1}{2}}}{\frac{1}{2}} + constante\\\\\\ \boxed{F_{(x)}=\dfrac{2x^{\frac{5}{2}}}{5}+2x^{\frac{1}{2}} + constante}$


Espero ter ajudado.
Bons estudos!