What is the integral of (5x^2-36x+48)/(x(x-4)^2) ?

The integral of (5x^2-36x+48)/(x(x-4)^2) is 2ln|x-4|+4/(x-4)+3ln|x|+C

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integral of (5x^2-36x+48)/(x(x-4)^2)

\int \frac{5 x ^{2} - 36 x + 48}{x ( x - 4 ) ^{2}} d x

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

Solution steps

\int \frac{5 x ^{2} - 36 x + 48}{x ( x - 4 ) ^{2}} d x

Expand

\frac{5 x ^{2} - 36 x + 48}{x ( x - 4 ) ^{2}} : \frac{5 x ^{2}}{x ( x - 4 ) ^{2}} - \frac{36 x}{x ( x - 4 ) ^{2}} + \frac{48}{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{48}{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{48}{x ( x - 4 ) ^{2}} d x = 48 (\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}) + 48 (\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}) + 48 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )}) : 2 ln ∣ x - 4 ∣ + \frac{4}{x - 4} + 3 ln ∣ x ∣

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

Add a constant to the solution

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

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What is the integral of (5x^2-36x+48)/(x(x-4)^2) ?
The integral of (5x^2-36x+48)/(x(x-4)^2) is 2ln|x-4|+4/(x-4)+3ln|x|+C