What is the integral of ((2x^3+10x^2+8x-12)/(4x)) ?

The integral of ((2x^3+10x^2+8x-12)/(4x)) is 1/4 ((2x^3)/3+5x^2+8x-12ln|x|)+C

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integral of ((2x^3+10x^2+8x-12)/(4x))

\int (\frac{2 x ^{3} + 10 x ^{2} + 8 x - 12}{4 x}) d x

\frac{1}{4} (\frac{2 x ^{3}}{3} + 5 x^{2} + 8 x - 12 ln ∣ x ∣) + C

Solution steps

\int \frac{2 x ^{3} + 10 x ^{2} + 8 x - 12}{4 x} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{2 x ^{3} + 10 x ^{2} + 8 x - 12}{x} d x

Expand

\frac{2 x ^{3} + 10 x ^{2} + 8 x - 12}{x} : 2 x^{2} + 10 x + 8 - \frac{12}{x}

= \frac{1}{4} \cdot \int 2 x^{2} + 10 x + 8 - \frac{12}{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}{4} (\int 2 x^{2} d x + \int 10 x d x + \int 8 d x - \int \frac{12}{x} d x)

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

\int 10 x d x = 5 x^{2}

\int 8 d x = 8 x

\int \frac{12}{x} d x = 12 ln ∣ x ∣

= \frac{1}{4} (\frac{2 x ^{3}}{3} + 5 x^{2} + 8 x - 12 ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{4} (\frac{2 x ^{3}}{3} + 5 x^{2} + 8 x - 12 ln ∣ x ∣) + C

\int \frac{x}{x ^{2} + 9} d x

y^{^{'''}} + 2 y^{^{''}} - 10 y^{^{'}} - 20 y = 0

f (x) = x^{2} + x - 7

x \to 2 lim (x^{2} - 3 x + 4)

\frac{d}{d x} (\frac{x ^{6}}{5 - x ^{5}})

What is the integral of ((2x^3+10x^2+8x-12)/(4x)) ?
The integral of ((2x^3+10x^2+8x-12)/(4x)) is 1/4 ((2x^3)/3+5x^2+8x-12ln|x|)+C