Evaluate the following integral. ∫(x+1x3)dx

Step 1
The given integral is $\int (\sqrt{x}+\frac{1}{\sqrt[3]{x}})dx$
Evaluate the above integral as follows.
Step 2
$\int (\sqrt{x}+\frac{1}{\sqrt[3]{x}})dx=\int (x^{\frac{1}{2}}+\frac{1}{3}{x}^{-\frac{1}{2}})dx$
${\displaystyle =\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{3}\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C}$
${\displaystyle =\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{1}{3}\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C}$
${\displaystyle =\frac{2}{3}{x}^{\frac{3}{2}}+\frac{2}{3}{x}^{\frac{1}{2}}+C}$
${\displaystyle =\frac{2}{3}(x\sqrt{x}+\sqrt{x})+C}$
Thus, $\int (\sqrt{x}+\frac{1}{\sqrt[3]{x}})dx=\frac{2}{3}(x\sqrt{x}+\sqrt{x})+C$

Step 1: Expand.
$\int (\sqrt{x}+\frac{1}{\sqrt[3]{x}})dx$
Step 2: Use Sum Rule: ${\displaystyle \int f\left(x\right)+g\left(x\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx}$.
${\displaystyle \int \sqrt{x}dx+\int \frac{1}{\sqrt{3}\left\{x\right\}}dx}$
Step 3: Since ${\displaystyle \sqrt{x}=x^{\frac{1}{2}}}$, using the PowerRule, ${\displaystyle \int x^{\frac{1}{2}}dx=\frac{2}{3}{x}^{\frac{3}{2}}}$
${\displaystyle \frac{2x^{\frac{3}{2}}}{3}+\int \frac{1}{\sqrt[3]{x}}dx}$
Step 4: Use Power Rule: ${\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C}$.
${\displaystyle \frac{2x^{\frac{3}{2}}}{3}+\frac{3}{2x^{-\frac{2}{3}}}}$
Step 5: Add constant.
${\displaystyle \frac{2x^{\frac{3}{2}}}{3}+\frac{3}{2x^{-\frac{2}{3}}}+C}$