What is the integral of (x^2-18x+32)/(x(x-4)^2) ?

The integral of (x^2-18x+32)/(x(x-4)^2) is -ln|x-4|+6/(x-4)+2ln|x|+C

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integral of (x^2-18x+32)/(x(x-4)^2)

\int \frac{x ^{2} - 18 x + 32}{x ( x - 4 ) ^{2}} d x

- ln ∣ x - 4 ∣ + \frac{6}{x - 4} + 2 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} - 18 x + 32}{x ( x - 4 ) ^{2}} d x

Expand

\frac{x ^{2} - 18 x + 32}{x ( x - 4 ) ^{2}} : \frac{x ^{2}}{x ( x - 4 ) ^{2}} - \frac{18 x}{x ( x - 4 ) ^{2}} + \frac{32}{x ( x - 4 ) ^{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 - 4 ) ^{2}} d x - \int \frac{18 x}{x ( x - 4 ) ^{2}} d x + \int \frac{32}{x ( x - 4 ) ^{2}} d x

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

\int \frac{18 x}{x ( x - 4 ) ^{2}} d x = - \frac{18}{x - 4}

\int \frac{32}{x ( x - 4 ) ^{2}} d x = 32 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )})

= ln ∣ x - 4 ∣ - \frac{4}{x - 4} - (- \frac{18}{x - 4}) + 32 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )})

Simplify

ln ∣ x - 4 ∣ - \frac{4}{x - 4} - (- \frac{18}{x - 4}) + 32 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )}) : - ln ∣ x - 4 ∣ + \frac{6}{x - 4} + 2 ln ∣ x ∣

= - ln ∣ x - 4 ∣ + \frac{6}{x - 4} + 2 ln ∣ x ∣

Add a constant to the solution

= - ln ∣ x - 4 ∣ + \frac{6}{x - 4} + 2 ln ∣ x ∣ + C

f (x) = x^{\frac{2}{5}}

2 sin (3 t) + 5 cos (3 t)

y^{^{''}} + 13 y = 0

\frac{d y}{d x} = \frac{x + x y ^{2}}{4 y}

\int_{0}^{\frac{3 π}{2}} cos^{3} (17 x) d x

What is the integral of (x^2-18x+32)/(x(x-4)^2) ?
The integral of (x^2-18x+32)/(x(x-4)^2) is -ln|x-4|+6/(x-4)+2ln|x|+C