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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

cos (2 x^{2})

\int_{0}^{2} \frac{3 ln ( x )}{x} d x

\frac{d}{d x} (\frac{1}{4} e^{x})

\frac{d}{d x} (ln (\frac{3 + x}{3 - x}))

\int_{0}^{1} (3 e^{x} - 3 x e^{x^{2}}) d x

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