What is the integral of (11w+4)/((3+6w-w^2)^2) ?

The integral of (11w+4)/((3+6w-w^2)^2) is

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integral of (11w+4)/((3+6w-w^2)^2)

\int \frac{11 w + 4}{( 3 + 6 w - w ^{2} ) ^{2}} d w

\frac{11}{2 ( - w ^{2} + 6 w + 3 )} + 37 (\frac{1}{96 3} ln w - 3 + 23 - \frac{1}{48 ( w - 3 + 2 3 )} - \frac{1}{96 3} ln w - 3 - 23 - \frac{1}{48 ( w - 3 - 2 3 )}) + C

Solution steps

\int \frac{11 w + 4}{( 3 + 6 w - w ^{2} ) ^{2}} d w

Complete the square

3 + 6 w - w^{2} : - (w - 3)^{2} + 12

= \int \frac{11 w + 4}{( - ( w - 3 ) ^{2} + 12 ) ^{2}} d w

Apply u-substitution

= \int \frac{11 u + 37}{( - u ^{2} + 12 ) ^{2}} d u

Expand

\frac{11 u + 37}{( - u ^{2} + 12 ) ^{2}} : \frac{11 u}{( - u ^{2} + 12 ) ^{2}} + \frac{37}{( - u ^{2} + 12 ) ^{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{11 u}{( - u ^{2} + 12 ) ^{2}} d u + \int \frac{37}{( - u ^{2} + 12 ) ^{2}} d u

\int \frac{11 u}{( - u ^{2} + 12 ) ^{2}} d u = \frac{11}{2 ( - u ^{2} + 12 )}

\int \frac{37}{( - u ^{2} + 12 ) ^{2}} d u = 37 (\frac{1}{96 3} ln u + 23 - \frac{1}{48 ( u + 2 3 )} - \frac{1}{96 3} ln u - 23 - \frac{1}{48 ( u - 2 3 )})

= \frac{11}{2 ( - u ^{2} + 12 )} + 37 (\frac{1}{96 3} ln u + 23 - \frac{1}{48 ( u + 2 3 )} - \frac{1}{96 3} ln u - 23 - \frac{1}{48 ( u - 2 3 )})

Substitute back

u = w - 3

= \frac{11}{2 ( - ( w - 3 ) ^{2} + 12 )} + 37 (\frac{1}{96 3} ln w - 3 + 23 - \frac{1}{48 ( w - 3 + 2 3 )} - \frac{1}{96 3} ln w - 3 - 23 - \frac{1}{48 ( w - 3 - 2 3 )})

Expand

- (w - 3)^{2} + 12 : - w^{2} + 6 w + 3

= \frac{11}{2 ( - w ^{2} + 6 w + 3 )} + 37 (\frac{1}{96 3} ln w - 3 + 23 - \frac{1}{48 ( w - 3 + 2 3 )} - \frac{1}{96 3} ln w - 3 - 23 - \frac{1}{48 ( w - 3 - 2 3 )})

Add a constant to the solution

= \frac{11}{2 ( - w ^{2} + 6 w + 3 )} + 37 (\frac{1}{96 3} ln w - 3 + 23 - \frac{1}{48 ( w - 3 + 2 3 )} - \frac{1}{96 3} ln w - 3 - 23 - \frac{1}{48 ( w - 3 - 2 3 )}) + C

What is the integral of (11w+4)/((3+6w-w^2)^2) ?
The integral of (11w+4)/((3+6w-w^2)^2) is