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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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}{u} d u - \int \frac{1}{u ^{2}} d u + \int \frac{1}{u - 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

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

Substitute back

u = x - 1

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

Simplify

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

Add a constant to the solution

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

y = 9 e^{x} cos (x)

1 - x

\int \frac{1}{m ^{3} + m} d m

\frac{d}{d x} (e^{- 6 x} - 6 e^{- 6 x} x)

x \to \infty lim (0.5 3^{2})

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