Question

∫ x+1 sobre √x dx alguem sabe passo a passo?

Answer 1

Não existe uma forma única de se resolver. Poderia separar em duas frações e ficar com duas integrais, mas vou fazer diferente:

$\displaystyle\int \dfrac{x+1}{\sqrt{x}}\,dx~~~~~~\mathbf{(i)}$


Forma 1:

Façamos a seguinte substituição:

$\sqrt{x}=u~~\Rightarrow~~\left\{ \!\begin{array}{l} x=u^2\\\\ dx=2u\,du \end{array} \right.$


e a integral $\mathbf{(i)}$ fica

$=\displaystyle\int \dfrac{u^2+1}{u}\cdot 2u\,du\\\\\\ =2\int (u^2+1)\,du\\\\\\ =2\cdot \left(\dfrac{u^3}{3}+u \right )+C\\\\\\ =\dfrac{2}{3}\,u^3+2u+C\\\\\\ =\dfrac{2}{3}\,(\sqrt{x})^3+2\sqrt{x}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int \dfrac{x+1}{\sqrt{x}}\,dx=\dfrac{2}{3}\,(\sqrt{x})^3+2\sqrt{x}+C \end{array}}$


Forma 2:

$\displaystyle\int \dfrac{x+1}{\sqrt{x}}\,dx\\\\\\ =\int \left(\dfrac{x}{\sqrt{x}}+\dfrac{1}{\sqrt{x}} \right )\,dx\\\\\\ =\int \left(\sqrt{x}+\dfrac{1}{\sqrt{x}} \right )\,dx\\\\\\ =\int \left(x^{1/2}+\dfrac{1}{x^{1/2}} \right )\,dx\\\\\\ =\int \big(x^{1/2}+x^{-1/2}\big)\,dx$

$=\dfrac{x^{(1/2)+1}}{\frac{1}{2}+1}+\dfrac{x^{(-1/2)+1}}{-\frac{1}{2}+1}+C\\\\\\ =\dfrac{x^{3/2}}{\frac{3}{2}}+\dfrac{x^{1/2}}{\frac{1}{2}}+C\\\\\\ =\dfrac{2}{3}\,x^{3/2}+2\,x^{1/2}+C\\\\\\ =\dfrac{2}{3}\,(\sqrt{x})^3+2\sqrt{x}+C\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int \dfrac{x+1}{\sqrt{x}}\,dx=\dfrac{2}{3}\,(\sqrt{x})^3+2\sqrt{x}+C \end{array}}$