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

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

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

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

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

Solution steps

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

Expand

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

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

\int 6 x d x = 3 x^{2}

\int 4 d x = 4 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

Simplify

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

= \frac{3}{2} x^{2} + 2 x - ln ∣ x ∣

Add a constant to the solution

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

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