What is the integral of ((x^3-12x+4))/((x^2-3x+2)^2) ?

The integral of ((x^3-12x+4))/((x^2-3x+2)^2) is -23ln|x-1|+7/(x-1)+24ln|x-2|+(12)/(x-2)+C

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integral of ((x^3-12x+4))/((x^2-3x+2)^2)

\int \frac{( x ^{3} - 12 x + 4 )}{( x ^{2} - 3 x + 2 ) ^{2}} d x

- 23 ln ∣ x - 1 ∣ + \frac{7}{x - 1} + 24 ln ∣ x - 2 ∣ + \frac{12}{x - 2} + C

Solution steps

\int \frac{x ^{3} - 12 x + 4}{( x ^{2} - 3 x + 2 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{3} - 12 x + 4}{( x ^{2} - 3 x + 2 ) ^{2}} : - \frac{23}{x - 1} - \frac{7}{( x - 1 ) ^{2}} + \frac{24}{x - 2} - \frac{12}{( x - 2 ) ^{2}}

= \int - \frac{23}{x - 1} - \frac{7}{( x - 1 ) ^{2}} + \frac{24}{x - 2} - \frac{12}{( x - 2 ) ^{2}} d x

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{23}{x - 1} d x - \int \frac{7}{( x - 1 ) ^{2}} d x + \int \frac{24}{x - 2} d x - \int \frac{12}{( x - 2 ) ^{2}} d x

\int \frac{23}{x - 1} d x = 23 ln ∣ x - 1 ∣

\int \frac{7}{( x - 1 ) ^{2}} d x = - \frac{7}{x - 1}

\int \frac{24}{x - 2} d x = 24 ln ∣ x - 2 ∣

\int \frac{12}{( x - 2 ) ^{2}} d x = - \frac{12}{x - 2}

= - 23 ln ∣ x - 1 ∣ - (- \frac{7}{x - 1}) + 24 ln ∣ x - 2 ∣ - (- \frac{12}{x - 2})

Simplify

= - 23 ln ∣ x - 1 ∣ + \frac{7}{x - 1} + 24 ln ∣ x - 2 ∣ + \frac{12}{x - 2}

Add a constant to the solution

= - 23 ln ∣ x - 1 ∣ + \frac{7}{x - 1} + 24 ln ∣ x - 2 ∣ + \frac{12}{x - 2} + C

y^{^{'}} = 5 y + y^{2}

\frac{d}{d x} (x^{2} cos (x^{3}))

(1.4) (2.2)

x \to 0 lim (\frac{∣ x - 2 ∣}{x - 2})

\frac{d ^{4}}{d x ^{4}} (1 + x)

What is the integral of ((x^3-12x+4))/((x^2-3x+2)^2) ?
The integral of ((x^3-12x+4))/((x^2-3x+2)^2) is -23ln|x-1|+7/(x-1)+24ln|x-2|+(12)/(x-2)+C