Find the Integral ( square root of x+ cube root of x)dx

$(\sqrt{x} + \sqrt[3]{x}) d x$

Step 1

Remove parentheses.

$\int \sqrt{x} + \sqrt[3]{x} d x$

Step 2

Split the single integral into multiple integrals.

$\int \sqrt{x} d x + \int \sqrt[3]{x} d x$

Step 3

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$\int x^{\frac{1}{2}} d x + \int \sqrt[3]{x} d x$

Step 4

By the Power Rule, the integral of $x^{\frac{1}{2}}$ with respect to $x$ is $\frac{2}{3} x^{\frac{3}{2}}$ .

$\frac{2}{3} x^{\frac{3}{2}} + C + \int \sqrt[3]{x} d x$

Step 5

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$ .

$\frac{2}{3} x^{\frac{3}{2}} + C + \int x^{\frac{1}{3}} d x$

Step 6

By the Power Rule, the integral of $x^{\frac{1}{3}}$ with respect to $x$ is $\frac{3}{4} x^{\frac{4}{3}}$ .

$\frac{2}{3} x^{\frac{3}{2}} + C + \frac{3}{4} x^{\frac{4}{3}} + C$

Step 7

Simplify.

$\frac{2}{3} x^{\frac{3}{2}} + \frac{3}{4} x^{\frac{4}{3}} + C$