What is the integral of (4x^3-6x+8)/(2x^2) ?

The integral of (4x^3-6x+8)/(2x^2) is x^2-3ln|x|-4/x+C

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integral of (4x^3-6x+8)/(2x^2)

\int \frac{4 x ^{3} - 6 x + 8}{2 x ^{2}} d x

x^{2} - 3 ln ∣ x ∣ - \frac{4}{x} + C

Solution steps

\int \frac{4 x ^{3} - 6 x + 8}{2 x ^{2}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{4 x ^{3} - 6 x + 8}{x ^{2}} d x

Expand

\frac{4 x ^{3} - 6 x + 8}{x ^{2}} : 4 x - \frac{6}{x} + \frac{8}{x ^{2}}

= \frac{1}{2} \cdot \int 4 x - \frac{6}{x} + \frac{8}{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}{2} (\int 4 x d x - \int \frac{6}{x} d x + \int \frac{8}{x ^{2}} d x)

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

\int \frac{6}{x} d x = 6 ln ∣ x ∣

\int \frac{8}{x ^{2}} d x = - \frac{8}{x}

= \frac{1}{2} (2 x^{2} - 6 ln ∣ x ∣ - \frac{8}{x})

Simplify

\frac{1}{2} (2 x^{2} - 6 ln ∣ x ∣ - \frac{8}{x}) : x^{2} - 3 ln ∣ x ∣ - \frac{4}{x}

= x^{2} - 3 ln ∣ x ∣ - \frac{4}{x}

Add a constant to the solution

= x^{2} - 3 ln ∣ x ∣ - \frac{4}{x} + C

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What is the integral of (4x^3-6x+8)/(2x^2) ?
The integral of (4x^3-6x+8)/(2x^2) is x^2-3ln|x|-4/x+C