What is the integral of (x^2+10x-36)/(x(x-3)^2) ?

The integral of (x^2+10x-36)/(x(x-3)^2) is 5ln|x-3|-1/(x-3)-4ln|x|+C

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integral of (x^2+10x-36)/(x(x-3)^2)

\int \frac{x ^{2} + 10 x - 36}{x ( x - 3 ) ^{2}} d x

5 ln ∣ x - 3 ∣ - \frac{1}{x - 3} - 4 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} + 10 x - 36}{x ( x - 3 ) ^{2}} d x

Expand

\frac{x ^{2} + 10 x - 36}{x ( x - 3 ) ^{2}} : \frac{x ^{2}}{x ( x - 3 ) ^{2}} + \frac{10 x}{x ( x - 3 ) ^{2}} - \frac{36}{x ( x - 3 ) ^{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 - 3 ) ^{2}} d x + \int \frac{10 x}{x ( x - 3 ) ^{2}} d x - \int \frac{36}{x ( x - 3 ) ^{2}} d x

\int \frac{x ^{2}}{x ( x - 3 ) ^{2}} d x = ln ∣ x - 3 ∣ - \frac{3}{x - 3}

\int \frac{10 x}{x ( x - 3 ) ^{2}} d x = - \frac{10}{x - 3}

\int \frac{36}{x ( x - 3 ) ^{2}} d x = 36 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{3 ( x - 3 )})

= ln ∣ x - 3 ∣ - \frac{3}{x - 3} - \frac{10}{x - 3} - 36 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{3 ( x - 3 )})

Simplify

ln ∣ x - 3 ∣ - \frac{3}{x - 3} - \frac{10}{x - 3} - 36 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{3 ( x - 3 )}) : 5 ln ∣ x - 3 ∣ - \frac{1}{x - 3} - 4 ln ∣ x ∣

= 5 ln ∣ x - 3 ∣ - \frac{1}{x - 3} - 4 ln ∣ x ∣

Add a constant to the solution

= 5 ln ∣ x - 3 ∣ - \frac{1}{x - 3} - 4 ln ∣ x ∣ + C

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What is the integral of (x^2+10x-36)/(x(x-3)^2) ?
The integral of (x^2+10x-36)/(x(x-3)^2) is 5ln|x-3|-1/(x-3)-4ln|x|+C