What is the integral of (x^2-3x-8)/(x^2-2x+1) ?

The integral of (x^2-3x-8)/(x^2-2x+1) is (x^3)/(x^2-2x+1)-ln|x-1|+7/(x-1)-1/((x-1)^2)+C

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integral of (x^2-3x-8)/(x^2-2x+1)

\int \frac{x ^{2} - 3 x - 8}{x ^{2} - 2 x + 1} d x

\frac{x ^{3}}{x ^{2} - 2 x + 1} - ln ∣ x - 1 ∣ + \frac{7}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

Solution steps

\int \frac{x ^{2} - 3 x - 8}{x ^{2} - 2 x + 1} d x

Expand

\frac{x ^{2} - 3 x - 8}{x ^{2} - 2 x + 1} : \frac{x ^{2}}{x ^{2} - 2 x + 1} - \frac{3 x}{x ^{2} - 2 x + 1} - \frac{8}{x ^{2} - 2 x + 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{x ^{2}}{x ^{2} - 2 x + 1} d x - \int \frac{3 x}{x ^{2} - 2 x + 1} d x - \int \frac{8}{x ^{2} - 2 x + 1} d x

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

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

\int \frac{8}{x ^{2} - 2 x + 1} d x = - \frac{8}{x - 1}

= \frac{x ^{3}}{x ^{2} - 2 x + 1} + 2 ln ∣ x - 1 ∣ - \frac{4}{x - 1} - \frac{1}{( x - 1 ) ^{2}} - 3 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - (- \frac{8}{x - 1})

Simplify

\frac{x ^{3}}{x ^{2} - 2 x + 1} + 2 ln ∣ x - 1 ∣ - \frac{4}{x - 1} - \frac{1}{( x - 1 ) ^{2}} - 3 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - (- \frac{8}{x - 1}) : \frac{x ^{3}}{x ^{2} - 2 x + 1} - ln ∣ x - 1 ∣ + \frac{7}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

= \frac{x ^{3}}{x ^{2} - 2 x + 1} - ln ∣ x - 1 ∣ + \frac{7}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

Add a constant to the solution

= \frac{x ^{3}}{x ^{2} - 2 x + 1} - ln ∣ x - 1 ∣ + \frac{7}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

x y - 6 y^{2} = 4, (14, 2)

\int x^{n - 1} a + b x^{n} d x

\frac{d}{d x} (x (x + 4)^{2})

\frac{\partial}{\partial x} (10 x^{2} y + 5 x y^{3})

\frac{d}{d x} ((ln (x))^{8})

What is the integral of (x^2-3x-8)/(x^2-2x+1) ?
The integral of (x^2-3x-8)/(x^2-2x+1) is (x^3)/(x^2-2x+1)-ln|x-1|+7/(x-1)-1/((x-1)^2)+C