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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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