What is the integral of (5x^2+6x+9)/((x-3)^2(x+1)^2) ?

The integral of (5x^2+6x+9)/((x-3)^2(x+1)^2) is -9/(2(x-3))-1/(2(x+1))+C

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integral of (5x^2+6x+9)/((x-3)^2(x+1)^2)

\int \frac{5 x ^{2} + 6 x + 9}{( x - 3 ) ^{2} ( x + 1 ) ^{2}} d x

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

Solution steps

\int \frac{5 x ^{2} + 6 x + 9}{( x - 3 ) ^{2} ( x + 1 ) ^{2}} d x

Take the partial fraction of

\frac{5 x ^{2} + 6 x + 9}{( x - 3 ) ^{2} ( x + 1 ) ^{2}} : \frac{9}{2 ( x - 3 ) ^{2}} + \frac{1}{2 ( x + 1 ) ^{2}}

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{6}}{1 + x ^{12}})

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

\frac{\partial}{\partial y} (x^{2} + y^{2} - 25)

\int - 0.04 t d t

\frac{d}{d x} (- \frac{2}{9} x^{- \frac{5}{3}})

What is the integral of (5x^2+6x+9)/((x-3)^2(x+1)^2) ?
The integral of (5x^2+6x+9)/((x-3)^2(x+1)^2) is -9/(2(x-3))-1/(2(x+1))+C