Question

A integral de f(x)= raiz de x (x + 1/x) = é?

Answer 1

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

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

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