What is the integral of (3x-18)/((x-9)(x-5)^2) ?

The integral of (3x-18)/((x-9)(x-5)^2) is 9/16 ln|x-9|-9/16 ln|x-5|+(15)/(4(x-5))-9/(2(x-5))+C

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integral of (3x-18)/((x-9)(x-5)^2)

\int \frac{3 x - 18}{( x - 9 ) ( x - 5 ) ^{2}} d x

\frac{9}{16} ln ∣ x - 9 ∣ - \frac{9}{16} ln ∣ x - 5 ∣ + \frac{15}{4 ( x - 5 )} - \frac{9}{2 ( x - 5 )} + C

Solution steps

\int \frac{3 x - 18}{( x - 9 ) ( x - 5 ) ^{2}} d x

Expand

\frac{3 x - 18}{( x - 9 ) ( x - 5 ) ^{2}} : \frac{3 x}{( x - 9 ) ( x - 5 ) ^{2}} - \frac{18}{( x - 9 ) ( x - 5 ) ^{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{3 x}{( x - 9 ) ( x - 5 ) ^{2}} d x - \int \frac{18}{( x - 9 ) ( x - 5 ) ^{2}} d x

\int \frac{3 x}{( x - 9 ) ( x - 5 ) ^{2}} d x = 3 (\frac{9}{16} ln ∣ x - 9 ∣ - \frac{9}{16} ln ∣ x - 5 ∣ + \frac{5}{4 ( x - 5 )})

\int \frac{18}{( x - 9 ) ( x - 5 ) ^{2}} d x = 18 (\frac{1}{16} ln ∣ x - 9 ∣ - \frac{1}{16} ln ∣ x - 5 ∣ + \frac{1}{4 ( x - 5 )})

= 3 (\frac{9}{16} ln ∣ x - 9 ∣ - \frac{9}{16} ln ∣ x - 5 ∣ + \frac{5}{4 ( x - 5 )}) - 18 (\frac{1}{16} ln ∣ x - 9 ∣ - \frac{1}{16} ln ∣ x - 5 ∣ + \frac{1}{4 ( x - 5 )})

Simplify

3 (\frac{9}{16} ln ∣ x - 9 ∣ - \frac{9}{16} ln ∣ x - 5 ∣ + \frac{5}{4 ( x - 5 )}) - 18 (\frac{1}{16} ln ∣ x - 9 ∣ - \frac{1}{16} ln ∣ x - 5 ∣ + \frac{1}{4 ( x - 5 )}) : \frac{9}{16} ln ∣ x - 9 ∣ - \frac{9}{16} ln ∣ x - 5 ∣ + \frac{15}{4 ( x - 5 )} - \frac{9}{2 ( x - 5 )}

= \frac{9}{16} ln ∣ x - 9 ∣ - \frac{9}{16} ln ∣ x - 5 ∣ + \frac{15}{4 ( x - 5 )} - \frac{9}{2 ( x - 5 )}

Add a constant to the solution

= \frac{9}{16} ln ∣ x - 9 ∣ - \frac{9}{16} ln ∣ x - 5 ∣ + \frac{15}{4 ( x - 5 )} - \frac{9}{2 ( x - 5 )} + C

\frac{d y}{d t} = 3 y + y^{3}

2 y^{^{''}} + 18 y = 5 cos (3 t)

\frac{d}{d x} (\frac{1}{( - x ^{2} + x - 2 ) ^{5}})

x^{2} y^{^{''}} + 5 x y^{^{'}} + 4 y = 0

\frac{d y}{d x} = \frac{1}{x ^{2}} - \frac{2 y}{x}

What is the integral of (3x-18)/((x-9)(x-5)^2) ?
The integral of (3x-18)/((x-9)(x-5)^2) is 9/16 ln|x-9|-9/16 ln|x-5|+(15)/(4(x-5))-9/(2(x-5))+C