What is the integral of (x^2-5x+16)/(x(x-2)^2) ?

The integral of (x^2-5x+16)/(x(x-2)^2) is -3ln|x-2|-5/(x-2)+4ln|x|+C

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integral of (x^2-5x+16)/(x(x-2)^2)

\int \frac{x ^{2} - 5 x + 16}{x ( x - 2 ) ^{2}} d x

- 3 ln ∣ x - 2 ∣ - \frac{5}{x - 2} + 4 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} - 5 x + 16}{x ( x - 2 ) ^{2}} d x

Expand

\frac{x ^{2} - 5 x + 16}{x ( x - 2 ) ^{2}} : \frac{x ^{2}}{x ( x - 2 ) ^{2}} - \frac{5 x}{x ( x - 2 ) ^{2}} + \frac{16}{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{5 x}{x ( x - 2 ) ^{2}} d x + \int \frac{16}{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{5 x}{x ( x - 2 ) ^{2}} d x = - \frac{5}{x - 2}

\int \frac{16}{x ( x - 2 ) ^{2}} d x = 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )})

= ln ∣ x - 2 ∣ - \frac{2}{x - 2} - (- \frac{5}{x - 2}) + 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )})

Simplify

ln ∣ x - 2 ∣ - \frac{2}{x - 2} - (- \frac{5}{x - 2}) + 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )}) : - 3 ln ∣ x - 2 ∣ - \frac{5}{x - 2} + 4 ln ∣ x ∣

= - 3 ln ∣ x - 2 ∣ - \frac{5}{x - 2} + 4 ln ∣ x ∣

Add a constant to the solution

= - 3 ln ∣ x - 2 ∣ - \frac{5}{x - 2} + 4 ln ∣ x ∣ + C

y = 12 x, y = x^{2}

n = 0 \sum \infty (2)^{n}

\frac{d}{d x} (\frac{x ^{3}}{3})

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

\int (x + 1)^{\frac{1}{3}} d x

What is the integral of (x^2-5x+16)/(x(x-2)^2) ?
The integral of (x^2-5x+16)/(x(x-2)^2) is -3ln|x-2|-5/(x-2)+4ln|x|+C