What is the integral of (sqrt(x)+1/(\sqrt[3]{x)})^2 ?

The integral of (sqrt(x)+1/(\sqrt[3]{x)})^2 is 3\sqrt[3]{x}+12/7 x^{7/6}+(x^2)/2+C

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integral of (sqrt(x)+1/(\sqrt[3]{x)})^2

\int (x + \frac{1}{3 x})^{2} d x

3 3 x + \frac{12}{7} x^{\frac{7}{6}} + \frac{x ^{2}}{2} + C

Solution steps

\int (x + \frac{1}{3 x})^{2} d x

Expand

(x + \frac{1}{3 x})^{2} : \frac{1}{x ^{\frac{2}{3}}} + 2 x^{\frac{1}{6}} + x

= \int \frac{1}{x ^{\frac{2}{3}}} + 2 x^{\frac{1}{6}} + x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x ^{\frac{2}{3}}} d x + \int 2 x^{\frac{1}{6}} d x + \int x d x

\int \frac{1}{x ^{\frac{2}{3}}} d x = 3 x^{\frac{1}{3}}

\int 2 x^{\frac{1}{6}} d x = \frac{12}{7} x^{\frac{7}{6}}

\int x d x = \frac{x ^{2}}{2}

= 3 x^{\frac{1}{3}} + \frac{12}{7} x^{\frac{7}{6}} + \frac{x ^{2}}{2}

Apply radical rule:

a^{\frac{1}{n}} = n a

= 3 3 x + \frac{12}{7} x^{\frac{7}{6}} + \frac{x ^{2}}{2}

Add a constant to the solution

= 3 3 x + \frac{12}{7} x^{\frac{7}{6}} + \frac{x ^{2}}{2} + C

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What is the integral of (sqrt(x)+1/(\sqrt[3]{x)})^2 ?
The integral of (sqrt(x)+1/(\sqrt[3]{x)})^2 is 3\sqrt[3]{x}+12/7 x^{7/6}+(x^2)/2+C