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

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

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integral of (x^3-4x^2+2x)/(x^2+1)

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

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

Solution steps

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

Expand

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

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

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

Simplify

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

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

Add a constant to the solution

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

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