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

The integral of (-x^2+x+4)/(x(x-2)^2) is -2ln|x-2|-1/(x-2)+ln|x|+C

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

\int \frac{- x ^{2} + x + 4}{x ( x - 2 ) ^{2}} d x

- 2 ln ∣ x - 2 ∣ - \frac{1}{x - 2} + ln ∣ x ∣ + C

Solution steps

\int \frac{- x ^{2} + x + 4}{x ( x - 2 ) ^{2}} d x

Expand

\frac{- x ^{2} + x + 4}{x ( x - 2 ) ^{2}} : - \frac{x ^{2}}{x ( x - 2 ) ^{2}} + \frac{x}{x ( x - 2 ) ^{2}} + \frac{4}{x ( x - 2 ) ^{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{x ^{2}}{x ( x - 2 ) ^{2}} d x + \int \frac{x}{x ( x - 2 ) ^{2}} d x + \int \frac{4}{x ( x - 2 ) ^{2}} d x

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

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

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

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

Simplify

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

= - 2 ln ∣ x - 2 ∣ - \frac{1}{x - 2} + ln ∣ x ∣

Add a constant to the solution

= - 2 ln ∣ x - 2 ∣ - \frac{1}{x - 2} + ln ∣ x ∣ + C

n = 1 \sum \infty ln (\frac{n + 1}{n})

9 y^{^{''}} - y = \frac{x e x}{3}

\int sin (x) x d x

\int_{3}^{8} x^{3} ln (3 x) d x

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 39 x e^{x}

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