What is the integral of (x/(x^2-5x+6))^2 ?

The integral of (x/(x^2-5x+6))^2 is 12ln|x-2|-4/(x-2)-12ln|x-3|-9/(x-3)+C

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integral of (x/(x^2-5x+6))^2

\int (\frac{x}{x ^{2} - 5 x + 6})^{2} d x

12 ln ∣ x - 2 ∣ - \frac{4}{x - 2} - 12 ln ∣ x - 3 ∣ - \frac{9}{x - 3} + C

Solution steps

\int (\frac{x}{x ^{2} - 5 x + 6})^{2} d x

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \int \frac{x ^{2}}{( x ^{2} - 5 x + 6 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x ^{2} - 5 x + 6 ) ^{2}} : \frac{12}{x - 2} + \frac{4}{( x - 2 ) ^{2}} - \frac{12}{x - 3} + \frac{9}{( x - 3 ) ^{2}}

= \int \frac{12}{x - 2} + \frac{4}{( x - 2 ) ^{2}} - \frac{12}{x - 3} + \frac{9}{( x - 3 ) ^{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{12}{x - 2} d x + \int \frac{4}{( x - 2 ) ^{2}} d x - \int \frac{12}{x - 3} d x + \int \frac{9}{( x - 3 ) ^{2}} d x

\int \frac{12}{x - 2} d x = 12 ln ∣ x - 2 ∣

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

\int \frac{12}{x - 3} d x = 12 ln ∣ x - 3 ∣

\int \frac{9}{( x - 3 ) ^{2}} d x = - \frac{9}{x - 3}

= 12 ln ∣ x - 2 ∣ - \frac{4}{x - 2} - 12 ln ∣ x - 3 ∣ - \frac{9}{x - 3}

Add a constant to the solution

= 12 ln ∣ x - 2 ∣ - \frac{4}{x - 2} - 12 ln ∣ x - 3 ∣ - \frac{9}{x - 3} + C

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What is the integral of (x/(x^2-5x+6))^2 ?
The integral of (x/(x^2-5x+6))^2 is 12ln|x-2|-4/(x-2)-12ln|x-3|-9/(x-3)+C