What is the integral of (-2x^2+8x-20)/(x^2-4x+8) ?

The integral of (-2x^2+8x-20)/(x^2-4x+8) is -4ln|x^2-4x+8|-2x+4ln| 1/4 x^2-x+2|-2arctan(1/2 x-1)+C

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integral of (-2x^2+8x-20)/(x^2-4x+8)

\int \frac{- 2 x ^{2} + 8 x - 20}{x ^{2} - 4 x + 8} d x

- 4 ln x^{2} - 4 x + 8 - 2 x + 4 ln \frac{1}{4} x^{2} - x + 2 - 2 arctan (\frac{1}{2} x - 1) + C

Solution steps

\int \frac{- 2 x ^{2} + 8 x - 20}{x ^{2} - 4 x + 8} d x

Expand

\frac{- 2 x ^{2} + 8 x - 20}{x ^{2} - 4 x + 8} : - \frac{2 x ^{2}}{x ^{2} - 4 x + 8} + \frac{8 x}{x ^{2} - 4 x + 8} - \frac{20}{x ^{2} - 4 x + 8}

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 x ^{2}}{x ^{2} - 4 x + 8} d x + \int \frac{8 x}{x ^{2} - 4 x + 8} d x - \int \frac{20}{x ^{2} - 4 x + 8} d x

\int \frac{2 x ^{2}}{x ^{2} - 4 x + 8} d x = 2 (2 ln x^{2} - 4 x + 8 + x)

\int \frac{8 x}{x ^{2} - 4 x + 8} d x = 8 (\frac{1}{2} ln \frac{1}{4} (x - 2)^{2} + 1 + arctan (\frac{1}{2} (x - 2)))

\int \frac{20}{x ^{2} - 4 x + 8} d x = 10 arctan (\frac{1}{2} (x - 2))

= - 2 (2 ln x^{2} - 4 x + 8 + x) + 8 (\frac{1}{2} ln \frac{1}{4} (x - 2)^{2} + 1 + arctan (\frac{1}{2} (x - 2))) - 10 arctan (\frac{1}{2} (x - 2))

Simplify

- 2 (2 ln x^{2} - 4 x + 8 + x) + 8 (\frac{1}{2} ln \frac{1}{4} (x - 2)^{2} + 1 + arctan (\frac{1}{2} (x - 2))) - 10 arctan (\frac{1}{2} (x - 2)) : - 4 ln x^{2} - 4 x + 8 - 2 x + 4 ln \frac{1}{4} x^{2} - x + 2 - 2 arctan (\frac{1}{2} x - 1)

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

Add a constant to the solution

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

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

x d y = 4 y d x

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

sin (x y) + π

\frac{d y}{d x} = \frac{x}{y 1 + x}

What is the integral of (-2x^2+8x-20)/(x^2-4x+8) ?
The integral of (-2x^2+8x-20)/(x^2-4x+8) is -4ln|x^2-4x+8|-2x+4ln| 1/4 x^2-x+2|-2arctan(1/2 x-1)+C