What is the integral of (-6x+18)/((x^2+9)^2) ?

The integral of (-6x+18)/((x^2+9)^2) is 3/(x^2+9)+1/3 (arctan(x/3)+1/2 sin(2arctan(x/3)))+C

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integral of (-6x+18)/((x^2+9)^2)

\int \frac{- 6 x + 18}{( x ^{2} + 9 ) ^{2}} d x

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

Solution steps

\int \frac{- 6 x + 18}{( x ^{2} + 9 ) ^{2}} d x

Expand

\frac{- 6 x + 18}{( x ^{2} + 9 ) ^{2}} : - \frac{6 x}{( x ^{2} + 9 ) ^{2}} + \frac{18}{( x ^{2} + 9 ) ^{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{6 x}{( x ^{2} + 9 ) ^{2}} d x + \int \frac{18}{( x ^{2} + 9 ) ^{2}} d x

\int \frac{6 x}{( x ^{2} + 9 ) ^{2}} d x = - \frac{3}{x ^{2} + 9}

\int \frac{18}{( x ^{2} + 9 ) ^{2}} d x = \frac{1}{3} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3})))

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

Simplify

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

Add a constant to the solution

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

What is the integral of (-6x+18)/((x^2+9)^2) ?
The integral of (-6x+18)/((x^2+9)^2) is 3/(x^2+9)+1/3 (arctan(x/3)+1/2 sin(2arctan(x/3)))+C