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

The integral of (x^4+\sqrt[3]{x}+3x)/(5x^3) is (5x^{14/3}-6x-30x^{5/3})/(50x^{8/3)}+C

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integral of (x^4+\sqrt[3]{x}+3x)/(5x^3)

\frac{5 x ^{\frac{14}{3}} - 6 x - 30 x ^{\frac{5}{3}}}{50 x ^{\frac{8}{3}}} + C

Solution steps

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Expand

= \frac{1}{5} \cdot \int x + \frac{1}{x ^{\frac{8}{3}}} + \frac{3}{x ^{2}} 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

= \frac{1}{5} (\int x d x + \int \frac{1}{x ^{\frac{8}{3}}} d x + \int \frac{3}{x ^{2}} d x)

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

\int \frac{1}{x ^{\frac{8}{3}}} d x = - \frac{3}{5 x ^{\frac{5}{3}}}

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

= \frac{1}{5} (\frac{x ^{2}}{2} - \frac{3}{5 x ^{\frac{5}{3}}} - \frac{3}{x})

Simplify

\frac{1}{5} (\frac{x ^{2}}{2} - \frac{3}{5 x ^{\frac{5}{3}}} - \frac{3}{x}) : \frac{5 x ^{\frac{14}{3}} - 6 x - 30 x ^{\frac{5}{3}}}{50 x ^{\frac{8}{3}}}

= \frac{5 x ^{\frac{14}{3}} - 6 x - 30 x ^{\frac{5}{3}}}{50 x ^{\frac{8}{3}}}

Add a constant to the solution

= \frac{5 x ^{\frac{14}{3}} - 6 x - 30 x ^{\frac{5}{3}}}{50 x ^{\frac{8}{3}}} + C

\frac{d}{d x} (- 8 ln (x))

y = 7 x^{- 7}

(3 x + 2 y + 2 y^{2}) d x + (x + 2 x y) d y = 0

\frac{d}{d t} (k^{2} sin^{2} (t))

2 csc^{3} (x) + 3 c t a (x^{2})

What is the integral of (x^4+\sqrt[3]{x}+3x)/(5x^3) ?
The integral of (x^4+\sqrt[3]{x}+3x)/(5x^3) is (5x^{14/3}-6x-30x^{5/3})/(50x^{8/3)}+C