Question

O povo aqui só sabe resolver integrais fáceis que nem eu mesmo ?

?$\int\limits{ \sqrt{ \frac{1+ \sqrt{x} }{x} } } \, dx$

Answer 1

$\int \sqrt{\frac{1 + \sqrt{x}}{x}} \ dx =$

 Consideremos $\sqrt{x} = \lambda \Rightarrow x = \lambda^2 \Rightarrow dx = 2\lambda \ d\lambda$.

 Então,

$\\ \int \sqrt{\frac{1 + \sqrt{x}}{x}} \ dx = \\\\\\ \int \sqrt{\frac{1 + \lambda}{\lambda^2}} \cdot 2\lambda \ d\lambda = \\\\\\ \int \frac{\sqrt{1 + \lambda}}{\lambda} \cdot 2\lambda \ d\lambda = \\\\\\ 2 \cdot \int \sqrt{1 + \lambda} \ d\lambda =$

 Façamos mais uma substituição simples!

 Seja $1 + \lambda = \delta$, então $d\lambda = d\delta$. Segue,

$\\ 2 \cdot \int \sqrt{1 + \lambda} \ d\lambda = \\\\\\ 2 \cdot \int \sqrt{\delta} \ d\delta = \\\\\\ 2 \cdot \int \delta^{\frac{1}{2}} \ d\delta = \\\\\\ 2 \cdot \left [ \frac{2}{3} \cdot \delta^{\frac{3}{2}} \right ] =$

 Voltando,

$\\ 2 \cdot \left [ \frac{2}{3} \cdot \delta^{\frac{3}{2}} \right ] = \\\\\\ \frac{4}{3} \cdot (1 + \lambda)^{\frac{3}{2}} + c = \\\\\\ \frac{4}{3} \cdot (1 + \sqrt{x})^{\frac{3}{2}} + c = \\\\\\ \boxed{\frac{4}{3} \cdot \sqrt{(1 + \sqrt{x})^3} + c}$