Question

A integral de f(x) = √x (x+1/x) é::

Answer 1

$\displaystyle\int{\sqrt{x}\left(x+\dfrac{1}{x} \right )dx}\\ \\ \\ =\int{x^{1/2}\cdot (x+x^{-1})\,dx}\\ \\ \\ =\int{(x^{1/2}\cdot x+x^{1/2}\cdot x^{-1})\,dx}\\ \\ \\ =\int{(x^{(1/2)+1}+x^{(1/2)-1})\,dx}\\ \\ \\ =\int{(x^{3/2}+x^{-1/2})\,dx}\\ \\ \\ =\int{x^{3/2}\,dx}+\int{x^{-1/2}\,dx}~~~~~~\mathbf{(i)}$

Regra da integral de potência:

$\displaystyle\int{x^{n}\,dx}=\dfrac{x^{n+1}}{n+1}\,,~~n\neq -1.$

Aplicando a regra acima em $\mathbf{(i)},$ temos

$\displaystyle\int{x^{3/2}\,dx}+\int{x^{-1/2}\,dx}\\ \\ \\ =\dfrac{x^{(3/2)+1}}{\frac{3}{2}+1}+\dfrac{x^{(-1/2)+1}}{-\frac{1}{2}+1}+C\\ \\ \\ =\dfrac{x^{5/2}}{\frac{5}{2}}+\dfrac{x^{1/2}}{\frac{1}{2}}+C\\ \\ \\ =\dfrac{2}{5}\,x^{5/2}+2x^{1/2}+C\\ \\ \\ \\ \Rightarrow\boxed{\begin{array}{c} \displaystyle\int{\sqrt{x}\left(x+\dfrac{1}{x} \right )dx}=\dfrac{2}{5}\,x^{5/2}+2x^{1/2}+C \end{array}}$