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

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

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

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

2 ln ∣ x - 3 ∣ - ln ∣ x - 2 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

= 2 ln ∣ u ∣ - ln ∣ u + 1 ∣

Substitute back

u = x - 3

= 2 ln ∣ x - 3 ∣ - ln ∣ x - 3 + 1 ∣

Simplify

= 2 ln ∣ x - 3 ∣ - ln ∣ x - 2 ∣

Add a constant to the solution

= 2 ln ∣ x - 3 ∣ - ln ∣ x - 2 ∣ + C

x \to 2 lim (∣ x ∣)

x \to 0 lim (x + 5)

\frac{d}{d x} (ln (3 x^{5} + 2))

n = 3 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}

\frac{d y}{d x} = 9 x 3 y + 7

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