What is the integral of (22x^3-44x^2+2)/(x^2-2x) ?

The integral of (22x^3-44x^2+2)/(x^2-2x) is 11x^2+ln|x-2|-ln|x|-88+C

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integral of (22x^3-44x^2+2)/(x^2-2x)

\int \frac{22 x ^{3} - 44 x ^{2} + 2}{x ^{2} - 2 x} d x

11 x^{2} + ln ∣ x - 2 ∣ - ln ∣ x ∣ - 88 + C

Solution steps

\int \frac{22 x ^{3} - 44 x ^{2} + 2}{x ^{2} - 2 x} d x

Complete the square

x^{2} - 2 x : (x - 1)^{2} - 1

= \int \frac{22 x ^{3} - 44 x ^{2} + 2}{( x - 1 ) ^{2} - 1} d x

Apply u-substitution

: \int \frac{22 ( u + 1 ) ^{3} - 44 ( u + 1 ) ^{2} + 2}{u ^{2} - 1} d u

= \int \frac{22 ( u + 1 ) ^{3} - 44 ( u + 1 ) ^{2} + 2}{u ^{2} - 1} d u

Expand

\frac{22 ( u + 1 ) ^{3} - 44 ( u + 1 ) ^{2} + 2}{u ^{2} - 1} : \frac{22 ( u + 1 ) ^{3}}{u ^{2} - 1} - \frac{44 ( u + 1 ) ^{2}}{u ^{2} - 1} + \frac{2}{u ^{2} - 1}

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{22 ( u + 1 ) ^{3}}{u ^{2} - 1} d u - \int \frac{44 ( u + 1 ) ^{2}}{u ^{2} - 1} d u + \int \frac{2}{u ^{2} - 1} d u

\int \frac{22 ( u + 1 ) ^{3}}{u ^{2} - 1} d u = 22 (\frac{1}{2} (u - 1)^{2} + 4 (u - 1) + 4 ln ∣ u - 1 ∣)

\int \frac{44 ( u + 1 ) ^{2}}{u ^{2} - 1} d u = 44 (2 ln ∣ u - 1 ∣ + u)

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

= 22 (\frac{1}{2} (u - 1)^{2} + 4 (u - 1) + 4 ln ∣ u - 1 ∣) - 44 (2 ln ∣ u - 1 ∣ + u) - ln ∣ u + 1 ∣ + ln ∣ u - 1 ∣

Substitute back

u = x - 1

= 22 (\frac{1}{2} (x - 1 - 1)^{2} + 4 (x - 1 - 1) + 4 ln ∣ x - 1 - 1 ∣) - 44 (2 ln ∣ x - 1 - 1 ∣ + x - 1) - ln ∣ x - 1 + 1 ∣ + ln ∣ x - 1 - 1 ∣

Simplify

22 (\frac{1}{2} (x - 1 - 1)^{2} + 4 (x - 1 - 1) + 4 ln ∣ x - 1 - 1 ∣) - 44 (2 ln ∣ x - 1 - 1 ∣ + x - 1) - ln ∣ x - 1 + 1 ∣ + ln ∣ x - 1 - 1 ∣ : 11 x^{2} + ln ∣ x - 2 ∣ - ln ∣ x ∣ - 88

= 11 x^{2} + ln ∣ x - 2 ∣ - ln ∣ x ∣ - 88

Add a constant to the solution

= 11 x^{2} + ln ∣ x - 2 ∣ - ln ∣ x ∣ - 88 + C

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What is the integral of (22x^3-44x^2+2)/(x^2-2x) ?
The integral of (22x^3-44x^2+2)/(x^2-2x) is 11x^2+ln|x-2|-ln|x|-88+C