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

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

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

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

\frac{1}{6} (5 ln ∣ 2 x - 1 ∣ - 2 ln ∣ 2 x - 4 ∣) + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 5}{2 u ( u - 3 )} d u

= \int \frac{u - 5}{2 u ( u - 3 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u - 5}{u ( u - 3 )} d u

Expand

\frac{u - 5}{u ( u - 3 )} : \frac{u}{u ( u - 3 )} - \frac{5}{u ( u - 3 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{u}{u ( u - 3 )} d u - \int \frac{5}{u ( u - 3 )} d u)

\int \frac{u}{u ( u - 3 )} d u = ln ∣ u - 3 ∣

\int \frac{5}{u ( u - 3 )} d u = 5 (- \frac{1}{3} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 3 ∣)

= \frac{1}{2} (ln ∣ u - 3 ∣ - 5 (- \frac{1}{3} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 3 ∣))

Substitute back

u = 2 x - 1

= \frac{1}{2} (ln ∣ 2 x - 1 - 3 ∣ - 5 (- \frac{1}{3} ln ∣ 2 x - 1 ∣ + \frac{1}{3} ln ∣ 2 x - 1 - 3 ∣))

Simplify

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

= \frac{1}{6} (5 ln ∣ 2 x - 1 ∣ - 2 ln ∣ 2 x - 4 ∣)

Add a constant to the solution

= \frac{1}{6} (5 ln ∣ 2 x - 1 ∣ - 2 ln ∣ 2 x - 4 ∣) + C

x \to \infty lim (x^{3} - 3)

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