What is the integral of (x^2-4x-4)/(x^3-2x^2+4x-8) ?

The integral of (x^2-4x-4)/(x^3-2x^2+4x-8) is -ln|x-2|+ln|x^2+4|+C

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integral of (x^2-4x-4)/(x^3-2x^2+4x-8)

\int \frac{x ^{2} - 4 x - 4}{x ^{3} - 2 x ^{2} + 4 x - 8} d x

- ln ∣ x - 2 ∣ + ln x^{2} + 4 + C

Solution steps

\int \frac{x ^{2} - 4 x - 4}{x ^{3} - 2 x ^{2} + 4 x - 8} d x

Take the partial fraction of

\frac{x ^{2} - 4 x - 4}{x ^{3} - 2 x ^{2} + 4 x - 8} : - \frac{1}{x - 2} + \frac{2 x}{x ^{2} + 4}

= \int - \frac{1}{x - 2} + \frac{2 x}{x ^{2} + 4} d x

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

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

\int \frac{2 x}{x ^{2} + 4} d x = ln x^{2} + 4

= - ln ∣ x - 2 ∣ + ln x^{2} + 4

Add a constant to the solution

= - ln ∣ x - 2 ∣ + ln x^{2} + 4 + C

What is the integral of (x^2-4x-4)/(x^3-2x^2+4x-8) ?
The integral of (x^2-4x-4)/(x^3-2x^2+4x-8) is -ln|x-2|+ln|x^2+4|+C