Question

Calcule a integral abaixo:
$\int\limits {( \sqrt{x} - \frac{1}{ \sqrt{x} } }) \, dx$

Answer 1

$\displaystyle I=\int\left(\sqrt x-\frac{1}{\sqrt x}\right)\,dx\\\\ I=\int\left(x^{\frac 1 2}-\frac{1}{x^{\frac 1 2}}\right)\,dx\\\\ I=\int\left(x^{\frac 1 2}-x^{-\frac 1 2}\right)\,dx$

$\displaystyle I=\int\left x^{\frac 1 2}\,dx-\int\left x^{-\frac 1 2}\,dx$

$\displaystyle I=\left[\dfrac{x^{\frac 1 2+1}}{\frac 1 2+1}\right]-\left[\dfrac{x^{-\frac 1 2+1}}{-\frac 1 2+1}\right]+C\\\\ I=\left[\dfrac{x^{\frac 3 2}}{\frac 3 2}\right]-\left[\dfrac{x^{\frac 1 2}}{\frac 1 2}\right]+C\\\\ I=\dfrac{2}{3}x^{\frac 3 2}-2x^{\frac 1 2}+C\\\\ \boxed{\int\left(\sqrt x-\frac{1}{\sqrt x}\right)\,dx=\dfrac{2}{3}x^{\frac 3 2}-2x^{\frac 1 2}+C}$