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

The integral of (x^2+1/(\sqrt[3]{x)})^2 is 3\sqrt[3]{x}+3/4 x^{8/3}+(x^5)/5+C

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

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

3 3 x + \frac{3}{4} x^{\frac{8}{3}} + \frac{x ^{5}}{5} + C

Solution steps

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

Expand

(x^{2} + \frac{1}{3 x})^{2} : \frac{1}{x ^{\frac{2}{3}}} + 2 x^{\frac{5}{3}} + x^{4}

= \int \frac{1}{x ^{\frac{2}{3}}} + 2 x^{\frac{5}{3}} + x^{4} 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{5}{3}} d x + \int x^{4} d x

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

\int 2 x^{\frac{5}{3}} d x = \frac{3}{4} x^{\frac{8}{3}}

\int x^{4} d x = \frac{x ^{5}}{5}

= 3 x^{\frac{1}{3}} + \frac{3}{4} x^{\frac{8}{3}} + \frac{x ^{5}}{5}

Apply radical rule:

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

= 3 3 x + \frac{3}{4} x^{\frac{8}{3}} + \frac{x ^{5}}{5}

Add a constant to the solution

= 3 3 x + \frac{3}{4} x^{\frac{8}{3}} + \frac{x ^{5}}{5} + C

y = 3 x e^{x}

x \to 1 + lim (x^{\frac{3}{1 - x}})

y^{^{'}} + y^{3} = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = (3 + x) e^{- 2 x}

- 4 e^{t}

What is the integral of (x^2+1/(\sqrt[3]{x)})^2 ?
The integral of (x^2+1/(\sqrt[3]{x)})^2 is 3\sqrt[3]{x}+3/4 x^{8/3}+(x^5)/5+C