What is the integral of (4x^2-13x+12)/(x(x-2)^2) ?

The integral of (4x^2-13x+12)/(x(x-2)^2) is ln|x-2|-1/(x-2)+3ln|x|+C

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integral of (4x^2-13x+12)/(x(x-2)^2)

\int \frac{4 x ^{2} - 13 x + 12}{x ( x - 2 ) ^{2}} d x

ln ∣ x - 2 ∣ - \frac{1}{x - 2} + 3 ln ∣ x ∣ + C

Solution steps

\int \frac{4 x ^{2} - 13 x + 12}{x ( x - 2 ) ^{2}} d x

Expand

\frac{4 x ^{2} - 13 x + 12}{x ( x - 2 ) ^{2}} : \frac{4 x ^{2}}{x ( x - 2 ) ^{2}} - \frac{13 x}{x ( x - 2 ) ^{2}} + \frac{12}{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{4 x ^{2}}{x ( x - 2 ) ^{2}} d x - \int \frac{13 x}{x ( x - 2 ) ^{2}} d x + \int \frac{12}{x ( x - 2 ) ^{2}} d x

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

\int \frac{13 x}{x ( x - 2 ) ^{2}} d x = - \frac{13}{x - 2}

\int \frac{12}{x ( x - 2 ) ^{2}} d x = 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )})

= 4 (ln ∣ x - 2 ∣ - \frac{2}{x - 2}) - (- \frac{13}{x - 2}) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )})

Simplify

4 (ln ∣ x - 2 ∣ - \frac{2}{x - 2}) - (- \frac{13}{x - 2}) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )}) : ln ∣ x - 2 ∣ - \frac{1}{x - 2} + 3 ln ∣ x ∣

= ln ∣ x - 2 ∣ - \frac{1}{x - 2} + 3 ln ∣ x ∣

Add a constant to the solution

= ln ∣ x - 2 ∣ - \frac{1}{x - 2} + 3 ln ∣ x ∣ + C

\int \frac{3 x ^{3}}{x ^{2} + 16} d x

\frac{d}{d x} (\frac{e ^{x}}{cos ( 3 x )})

\int cot^{2} (x) csc^{2} (x) d x

\int 7 x^{3} e^{x^{4}} d x

\frac{d}{d x} ((4 x^{2} - 5)^{3})

What is the integral of (4x^2-13x+12)/(x(x-2)^2) ?
The integral of (4x^2-13x+12)/(x(x-2)^2) is ln|x-2|-1/(x-2)+3ln|x|+C