Question

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

Answer 1

$I=\displaystyle\int\sqrt x\left(x+\frac 1 x\right)\,dx\\\\ I=\int x^{\frac 1 2}\left(x+\frac 1 x\right)\,dx\\\\ I=\int x^{\frac 1 2}\cdot x\,dx+\int x^{\frac 1 2}\cdot \frac 1 x\,dx\\\\ I=\int x^{\frac 1 2+1}\,dx+\int x^{\frac 1 2}\cdot x^{-1}\,dx\\\\ I=\int x^{\frac 3 2}\,dx+\int x^{-\frac 1 2}\,dx\\\\ I=\left[\dfrac{x^{\frac 3 2+1}}{\frac 3 2+1}\right]+\left[\dfrac{x^{-\frac 1 2+1}}{-\frac 1 2+1}\right]+C\\\\ I=\left[\dfrac{x^{\frac 5 2}}{\frac 5 2}\right]+\left[\dfrac{x^{\frac 1 2}}{\frac 1 2}\right]+C$

$I=\dfrac{2}{5}x^{\frac 5 2}+2x^{\frac 1 2}+C\\\\ \boxed{\displaystyle\int\sqrt x\left(x+\frac 1 x\right)\,dx=\dfrac{2}{5}x^{\frac 5 2}+2x^{\frac 1 2}+C}$