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

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

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

\int \frac{x ^{2} - x + 4}{( x ^{2} + 4 ) ^{2}} d x

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

Solution steps

\int \frac{x ^{2} - x + 4}{( x ^{2} + 4 ) ^{2}} d x

Take the partial fraction of

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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