What is the integral of (3x+8)/(x(x-2)^2) ?

The integral of (3x+8)/(x(x-2)^2) is -7/(x-2)+2ln|x|-2ln|x-2|+C

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integral of (3x+8)/(x(x-2)^2)

\int \frac{3 x + 8}{x ( x - 2 ) ^{2}} d x

- \frac{7}{x - 2} + 2 ln ∣ x ∣ - 2 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{3 x + 8}{x ( x - 2 ) ^{2}} d x

Expand

\frac{3 x + 8}{x ( x - 2 ) ^{2}} : \frac{3 x}{x ( x - 2 ) ^{2}} + \frac{8}{x ( x - 2 ) ^{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 ( x - 2 ) ^{2}} d x + \int \frac{8}{x ( x - 2 ) ^{2}} d x

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

\int \frac{8}{x ( x - 2 ) ^{2}} d x = 8 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )})

= - \frac{3}{x - 2} + 8 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )})

Simplify

- \frac{3}{x - 2} + 8 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )}) : - \frac{7}{x - 2} + 2 ln ∣ x ∣ - 2 ln ∣ x - 2 ∣

= - \frac{7}{x - 2} + 2 ln ∣ x ∣ - 2 ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{7}{x - 2} + 2 ln ∣ x ∣ - 2 ln ∣ x - 2 ∣ + C

What is the integral of (3x+8)/(x(x-2)^2) ?
The integral of (3x+8)/(x(x-2)^2) is -7/(x-2)+2ln|x|-2ln|x-2|+C