What is the integral of (20x^4-3x^2+15x)/(5x^2) ?

The integral of (20x^4-3x^2+15x)/(5x^2) is 1/5 ((20x^3)/3-3x+15ln|x|)+C

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integral of (20x^4-3x^2+15x)/(5x^2)

\int \frac{20 x ^{4} - 3 x ^{2} + 15 x}{5 x ^{2}} d x

\frac{1}{5} (\frac{20 x ^{3}}{3} - 3 x + 15 ln ∣ x ∣) + C

Solution steps

\int \frac{20 x ^{4} - 3 x ^{2} + 15 x}{5 x ^{2}} d x

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{20 x ^{4} - 3 x ^{2} + 15 x}{x ^{2}} d x

Simplify

\frac{20 x ^{4} - 3 x ^{2} + 15 x}{x ^{2}} : \frac{20 x ^{3} - 3 x + 15}{x}

= \frac{1}{5} \cdot \int \frac{20 x ^{3} - 3 x + 15}{x} d x

Expand

\frac{20 x ^{3} - 3 x + 15}{x} : 20 x^{2} - 3 + \frac{15}{x}

= \frac{1}{5} \cdot \int 20 x^{2} - 3 + \frac{15}{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

= \frac{1}{5} (\int 20 x^{2} d x - \int 3 d x + \int \frac{15}{x} d x)

\int 20 x^{2} d x = \frac{20 x ^{3}}{3}

\int 3 d x = 3 x

\int \frac{15}{x} d x = 15 ln ∣ x ∣

= \frac{1}{5} (\frac{20 x ^{3}}{3} - 3 x + 15 ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{5} (\frac{20 x ^{3}}{3} - 3 x + 15 ln ∣ x ∣) + C

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What is the integral of (20x^4-3x^2+15x)/(5x^2) ?
The integral of (20x^4-3x^2+15x)/(5x^2) is 1/5 ((20x^3)/3-3x+15ln|x|)+C