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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( u + 1 ) ^{2}} d u

= \int \frac{1}{u ( u + 1 ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{( v - 1 ) v ^{2}} d v

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

Take the partial fraction of

\frac{1}{( v - 1 ) v ^{2}} : \frac{1}{v - 1} - \frac{1}{v} - \frac{1}{v ^{2}}

= \int \frac{1}{v - 1} - \frac{1}{v} - \frac{1}{v ^{2}} d v

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}{v - 1} d v - \int \frac{1}{v} d v - \int \frac{1}{v ^{2}} d v

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

= ln ∣ v - 1 ∣ - ln ∣ v ∣ - (- \frac{1}{v})

Substitute back

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

Simplify

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

Add a constant to the solution

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

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