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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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