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

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

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

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

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

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{2 x ^{2}}{x ^{2} ( x + 2 )} d x = 2 ln ∣ x + 2 ∣

\int \frac{6 x}{x ^{2} ( x + 2 )} d x = 6 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣)

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

= - 2 ln ∣ x + 2 ∣ + 6 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣) + 8 (- \frac{1}{4} ln ∣ x ∣ - \frac{1}{2 x} + \frac{1}{4} ln ∣ x + 2 ∣)

Simplify

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

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

Add a constant to the solution

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

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