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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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