What is the integral of 1/((x-2)^2(x-3)) ?

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

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integral of 1/((x-2)^2(x-3))

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

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

Solution steps

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

Take the partial fraction of

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

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

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

Simplify

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

Add a constant to the solution

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

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