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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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