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

The integral of ((2x^4+2\sqrt[5]{x}))/x is (x^4)/2+10\sqrt[5]{x}+C

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integral of ((2x^4+2\sqrt[5]{x}))/x

\int \frac{( 2 x ^{4} + 2 5 x )}{x} d x

\frac{x ^{4}}{2} + 10 5 x + C

Solution steps

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

Expand

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

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

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

\int \frac{2}{x ^{\frac{4}{5}}} d x = 10 x^{\frac{1}{5}}

= \frac{x ^{4}}{2} + 10 x^{\frac{1}{5}}

Apply radical rule:

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

= \frac{x ^{4}}{2} + 10 5 x

Add a constant to the solution

= \frac{x ^{4}}{2} + 10 5 x + C

t \to 0 lim (\frac{( 2 + t ) ^{3} - 8}{t})

x \to - 2 lim (4 x + 1)

\frac{d y}{d t} = - 5 (y - 2)

(x + 1)^{5}

y = \frac{2}{3} x^{\frac{3}{2}}

What is the integral of ((2x^4+2\sqrt[5]{x}))/x ?
The integral of ((2x^4+2\sqrt[5]{x}))/x is (x^4)/2+10\sqrt[5]{x}+C