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

The integral of (x^2-x-2)/(x^3-2x-4) is 1/2 ln|x^2+2x+2|+C

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

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

\frac{1}{2} ln x^{2} + 2 x + 2 + C

Solution steps

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

Take the partial fraction of

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

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

Expand

\frac{x + 1}{x ^{2} + 2 x + 2} : \frac{x}{x ^{2} + 2 x + 2} + \frac{1}{x ^{2} + 2 x + 2}

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

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

\int \frac{1}{x ^{2} + 2 x + 2} d x = arctan (x + 1)

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

Simplify

= \frac{1}{2} ln x^{2} + 2 x + 2

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 2 x + 2 + C

4 cos^{4} (x)

y = (2 x^{2} + 5) (4 x - 5)

\int \frac{( x ^{2} + 11 x - 30 )}{x ^{3} - 5 x ^{2} + 6 x} d x

\frac{d y}{d x}, y = cos (x - y)

\int - 6 e^{- 4 c s c (3 x)} cot (3 x) csc (3 x) d x

What is the integral of (x^2-x-2)/(x^3-2x-4) ?
The integral of (x^2-x-2)/(x^3-2x-4) is 1/2 ln|x^2+2x+2|+C