zachutnat4o

2021-11-22

Evaluate the following integral.
$\int \left(\sqrt{x}+\frac{1}{\sqrt[3]{x}}\right)dx$

Warajected53

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

Harr1957

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