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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

\int - csc^{2} (\frac{7 x}{6}) d x

\int_{0}^{5} \frac{8 t}{( t - 6 ) ^{2}} d t

\int x^{\frac{2}{5}} d x

x \to - 2 lim (e^{2 x - x^{2}})

\frac{d}{d x} (- 2 x^{- 2})

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