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

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

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

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

\frac{x ^{2}}{2} - 6 ln ∣ x ∣ - \frac{8}{x} + C

Solution steps

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

Expand

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

= \int 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

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

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

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

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

Add a constant to the solution

= \frac{x ^{2}}{2} - 6 ln ∣ x ∣ - \frac{8}{x} + C

\int \frac{2}{( x ^{2} - 1 )} d x

(e^{4 x})^{^{'}}

\frac{d}{d x} (5 \cdot e^{x} + 2 \cdot 3^{x})

e^{- t^{2}}

\int (x^{2} e^{x}) d x

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