What is the integral of (x^2-x+9)/((x^2+9)^2) ?

The integral of (x^2-x+9)/((x^2+9)^2) is 1/3 arctan(x/3)+1/(2(x^2+9))+C

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integral of (x^2-x+9)/((x^2+9)^2)

\int \frac{x ^{2} - x + 9}{( x ^{2} + 9 ) ^{2}} d x

\frac{1}{3} arctan (\frac{x}{3}) + \frac{1}{2 ( x ^{2} + 9 )} + C

Solution steps

\int \frac{x ^{2} - x + 9}{( x ^{2} + 9 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} - x + 9}{( x ^{2} + 9 ) ^{2}} : \frac{1}{x ^{2} + 9} - \frac{x}{( x ^{2} + 9 ) ^{2}}

= \int \frac{1}{x ^{2} + 9} - \frac{x}{( x ^{2} + 9 ) ^{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} + 9} d x - \int \frac{x}{( x ^{2} + 9 ) ^{2}} d x

\int \frac{1}{x ^{2} + 9} d x = \frac{1}{3} arctan (\frac{x}{3})

\int \frac{x}{( x ^{2} + 9 ) ^{2}} d x = - \frac{1}{2 ( x ^{2} + 9 )}

= \frac{1}{3} arctan (\frac{x}{3}) - (- \frac{1}{2 ( x ^{2} + 9 )})

Simplify

= \frac{1}{3} arctan (\frac{x}{3}) + \frac{1}{2 ( x ^{2} + 9 )}

Add a constant to the solution

= \frac{1}{3} arctan (\frac{x}{3}) + \frac{1}{2 ( x ^{2} + 9 )} + C

\int cos (2 x) 2 d x

x \to - 5 + lim (\frac{x - 5}{x + 5})

\frac{d x}{d t} = 375 - \frac{3 x}{800 + 2 t}

\frac{x ^{4}}{12}

y^{^{'}} = \frac{( y ^{2} + x y - x ^{2} )}{x ^{2}}

What is the integral of (x^2-x+9)/((x^2+9)^2) ?
The integral of (x^2-x+9)/((x^2+9)^2) is 1/3 arctan(x/3)+1/(2(x^2+9))+C