What is the integral of ((5x^2-36x+18)/(x(x-4)^2)) ?

The integral of ((5x^2-36x+18)/(x(x-4)^2)) is 9/8 ln|x|+31/8 ln|x-4|+(16)/(x-4)-9/(2(x-4))+C

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integral of ((5x^2-36x+18)/(x(x-4)^2))

\int (\frac{5 x ^{2} - 36 x + 18}{x ( x - 4 ) ^{2}}) d x

\frac{9}{8} ln ∣ x ∣ + \frac{31}{8} ln ∣ x - 4 ∣ + \frac{16}{x - 4} - \frac{9}{2 ( x - 4 )} + C

Solution steps

\int \frac{5 x ^{2} - 36 x + 18}{x ( x - 4 ) ^{2}} d x

Expand

\frac{5 x ^{2} - 36 x + 18}{x ( x - 4 ) ^{2}} : \frac{5 x ^{2}}{x ( x - 4 ) ^{2}} - \frac{36 x}{x ( x - 4 ) ^{2}} + \frac{18}{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{5 x ^{2}}{x ( x - 4 ) ^{2}} d x - \int \frac{36 x}{x ( x - 4 ) ^{2}} d x + \int \frac{18}{x ( x - 4 ) ^{2}} d x

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

\int \frac{36 x}{x ( x - 4 ) ^{2}} d x = - \frac{36}{x - 4}

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

= 5 (ln ∣ x - 4 ∣ - \frac{4}{x - 4}) - (- \frac{36}{x - 4}) + 18 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )})

Simplify

5 (ln ∣ x - 4 ∣ - \frac{4}{x - 4}) - (- \frac{36}{x - 4}) + 18 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )}) : \frac{9}{8} ln ∣ x ∣ + \frac{31}{8} ln ∣ x - 4 ∣ + \frac{16}{x - 4} - \frac{9}{2 ( x - 4 )}

= \frac{9}{8} ln ∣ x ∣ + \frac{31}{8} ln ∣ x - 4 ∣ + \frac{16}{x - 4} - \frac{9}{2 ( x - 4 )}

Add a constant to the solution

= \frac{9}{8} ln ∣ x ∣ + \frac{31}{8} ln ∣ x - 4 ∣ + \frac{16}{x - 4} - \frac{9}{2 ( x - 4 )} + C

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What is the integral of ((5x^2-36x+18)/(x(x-4)^2)) ?
The integral of ((5x^2-36x+18)/(x(x-4)^2)) is 9/8 ln|x|+31/8 ln|x-4|+(16)/(x-4)-9/(2(x-4))+C