What is the integral of 1/((x^2-4x+4)(x^2-5x+6)) ?

The integral of 1/((x^2-4x+4)(x^2-5x+6)) is -ln|x-2|+1/(x-2)+1/(2(x-2)^2)+ln|x-3|+C

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integral of 1/((x^2-4x+4)(x^2-5x+6))

\int \frac{1}{( x ^{2} - 4 x + 4 ) ( x ^{2} - 5 x + 6 )} d x

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

Solution steps

\int \frac{1}{( x ^{2} - 4 x + 4 ) ( x ^{2} - 5 x + 6 )} d x

Take the partial fraction of

\frac{1}{( x ^{2} - 4 x + 4 ) ( x ^{2} - 5 x + 6 )} : - \frac{1}{x - 2} - \frac{1}{( x - 2 ) ^{2}} - \frac{1}{( x - 2 ) ^{3}} + \frac{1}{x - 3}

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

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

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

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

\int \frac{1}{x - 3} d x = ln ∣ x - 3 ∣

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

Simplify

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

Add a constant to the solution

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

\int - 1 sin^{3} (x) cos^{2} (x) d x

a re a x^{2} + 1, 3 - x, x = 0

\frac{d}{d x} ((\frac{tan ( x )}{x})^{2})

n = 1 \sum \infty n ln (2)

\int sin^{2} (v) cos^{3} (v) d v

What is the integral of 1/((x^2-4x+4)(x^2-5x+6)) ?
The integral of 1/((x^2-4x+4)(x^2-5x+6)) is -ln|x-2|+1/(x-2)+1/(2(x-2)^2)+ln|x-3|+C