What is the integral of (x(2x-9))/(x^3-6x^2+12x-8) ?

The integral of (x(2x-9))/(x^3-6x^2+12x-8) is 2ln|x-2|+1/(x-2)+5/((x-2)^2)+C

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integral of (x(2x-9))/(x^3-6x^2+12x-8)

\int \frac{x ( 2 x - 9 )}{x ^{3} - 6 x ^{2} + 12 x - 8} d x

2 ln ∣ x - 2 ∣ + \frac{1}{x - 2} + \frac{5}{( x - 2 ) ^{2}} + C

Solution steps

\int \frac{x ( 2 x - 9 )}{x ^{3} - 6 x ^{2} + 12 x - 8} d x

Take the partial fraction of

\frac{x ( 2 x - 9 )}{x ^{3} - 6 x ^{2} + 12 x - 8} : \frac{2}{x - 2} - \frac{1}{( x - 2 ) ^{2}} - \frac{10}{( x - 2 ) ^{3}}

= \int \frac{2}{x - 2} - \frac{1}{( x - 2 ) ^{2}} - \frac{10}{( x - 2 ) ^{3}} 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{2}{x - 2} d x - \int \frac{1}{( x - 2 ) ^{2}} d x - \int \frac{10}{( x - 2 ) ^{3}} d x

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

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

\int \frac{10}{( x - 2 ) ^{3}} d x = - \frac{5}{( x - 2 ) ^{2}}

= 2 ln ∣ x - 2 ∣ - (- \frac{1}{x - 2}) - (- \frac{5}{( x - 2 ) ^{2}})

Simplify

= 2 ln ∣ x - 2 ∣ + \frac{1}{x - 2} + \frac{5}{( x - 2 ) ^{2}}

Add a constant to the solution

= 2 ln ∣ x - 2 ∣ + \frac{1}{x - 2} + \frac{5}{( x - 2 ) ^{2}} + C

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What is the integral of (x(2x-9))/(x^3-6x^2+12x-8) ?
The integral of (x(2x-9))/(x^3-6x^2+12x-8) is 2ln|x-2|+1/(x-2)+5/((x-2)^2)+C