Question

Integral de 2- raiz de x/ raiz de x dx

Answer 1

$I=\displaystyle\int{\dfrac{2-\sqrt{x}}{\sqrt{x}}\,dx}\\ \\ \\ =-2\int{-\dfrac{2-\sqrt{x}}{2\sqrt{x}}\,dx}\\ \\ \\ =-2\int{(2-\sqrt{x})\cdot\left(-\dfrac{1}{2\sqrt{x}} \right )dx}~~~~~~\mathbf{(i)}$


Façamos a seguinte substituição:

$2-\sqrt{x}=u~~\Rightarrow~~-\dfrac{1}{2\sqrt{x}}\,dx=du$


Substituindo em $\mathbf{(i)},$ a integral fica

$=\displaystyle-2\int{u\,du}\\ \\ \\ =-2\cdot \dfrac{u^{2}}{2}+C\\ \\ \\ =-\diagup\!\!\!\! 2\cdot \dfrac{u^{2}}{\diagup\!\!\!\! 2}+C\\ \\ \\ =-u^{2}+C\\ \\ \\ =-(2-\sqrt{x})^{2}+C\\ \\ \\ \\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int{\dfrac{2-\sqrt{x}}{\sqrt{x}}\,dx}=-(2-\sqrt{x})^{2}+C \end{array}}$

Answer 2

Resolvendo a integral por partes:

$\int u.dv = u.v-\int v.du\\\\\\\\ F_{(x)}= {\displaystyle \int \dfrac{2-\sqrt{x}}{\sqrt{x}}.dx\\\\ F_{(x)}= {\displaystyle \int (2-\sqrt{x})\dfrac{1}{\sqrt{x}}.dx\\\\ F_{(x)}= {\displaystyle \int (\underbrace{2-\sqrt{x}}) (\underbrace{x^{-\frac{1}{2}}.dx})\\ \ ~~~~~~~~~~~~~~~~~u~~~~~~~~~~dv$


$u=2-\sqrt{x}\ \ \ \ \ \ \ \ \ du=-\dfrac{1}{2\sqrt{x}}.dx\\\\ dv=x^{-\frac{1}{2}}.dx\ \ \ \ \ \ \ \ v=2\sqrt{x}\\\\\\\\ \displaystyle \int (2-\sqrt{x})(x^{-\frac{1}{2}}dx)=(2-\sqrt{x}).(2\sqrt{x})-\int (2\sqrt{x})\left(-\dfrac{1}{2\sqrt{x}}\right)dx\\\\ \displaystyle \int (2-\sqrt{x})(x^{-\frac{1}{2}}dx)=4\sqrt{x}-2x+\int \left(\dfrac{2\sqrt{x}}{2\sqrt{x}}\right)dx\\\\ \displaystyle \int (2-\sqrt{x})(x^{-\frac{1}{2}}dx)=4\sqrt{x}-2x+\int 1dx$

$\displaystyle \int (2-\sqrt{x})(x^{-\frac{1}{2}}dx)=4\sqrt{x}-2x+x+constante\\\\ \displaystyle \int (2-\sqrt{x})(x^{-\frac{1}{2}}dx)=4\sqrt{x}-x+constante\\\\\\ \boxed{F_{(x)}=4\sqrt{x}-x+constante}$



Espero ter ajudado.
Bons estudos!