What is the integral of (2x^3+5x)/(3x^5) ?

The integral of (2x^3+5x)/(3x^5) is 1/3 (-2/x-5/(3x^3))+C

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integral of (2x^3+5x)/(3x^5)

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

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

Solution steps

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

Take the constant out:

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

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

Simplify

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

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

Expand

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

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

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

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

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

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

Add a constant to the solution

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

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What is the integral of (2x^3+5x)/(3x^5) ?
The integral of (2x^3+5x)/(3x^5) is 1/3 (-2/x-5/(3x^3))+C