Question

integrate the following
$( \sqrt{x \: } + \: \frac{1}{3 \sqrt{x} } )dx$
​

Answer 1

Answer:

The answer is $\frac{2}{3}\sqrt x(x+1)+c$

Step-by-step explanation:

      $\displaystyle \int (\sqrt x+\frac{1}{3\sqrt x})dx$

   $=\displaystyle \int(x^{\frac{1}{2}}+\frac{1}{3}x^{-\frac{1}{2}})dx\\=\displaystyle \int x^{\frac{1}{2}}dx+\frac{1}{3}\int x^{-\frac{1}{2}}dx\\\\=\dfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{3}\dfrac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c$

• Since $\displaystyle \int x^ndx =\frac{x^{n+1}}{n+1}+C$

      $=\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{1}{3}\dfrac{x^{\frac{1}{2}}}{\frac{1}{2}}+c\\\\=\frac{2}{3}x^{\frac{3}{2}}+\frac{2}{3}x^{\frac{1}{2}}+c\\\\=\frac{2}{3}\sqrt x(x+1)+c$