How do you integrate #1/(9x^2+4)^2 dx#?

Let, #I=int1/(9x^2+4)^2dx#.

#9x^2+4=(3x)^2+2^2#, suggests that, #3x=2tany# substitution

may work.

So, let, #3x=2tany. :. 3dx=2sec^2ydy#.

#:. I=int1/(4tan^2y+4)^2*2/3sec^2ydy#,

#=2/48intsec^2y/sec^4ydy#,

#=1/24intcos^2ydy#,

#=1/24int(1+cos2y)/2dy#,

#=1/48{y+(sin2y)/2}#,

#=1/48{y+1/2*(2tany)/(1+tan^2y)}#,

Here, #3x=2tany rArr y=arc tan((3x)/2)#.

#:. I=1/48{arc tan((3x)/2)+((3x)/2)/(1+((3x)/2)^2)}#.

# rArr I=1/48{arc tan((3x)/2)+(6x)/(9x^2+4)}+C#.

As a first simplification, let's take the 9 out of the denominator:

# int \frac{1}{(9 x^2 + 4)^2} dx = \frac{1}{81} int \frac{1}{(x^2 + \frac{4}{9})^2} dx #.

If we have a look at the denominator now, then it somehow reminds us of the (RHS of the) trigonometric identity #sec^2 x = tan^2 x + 1#, and besides this, we remember that #(tan x)' = sec^2 x#. Therefore, we make the substitution #x \mapsto g ( t )# with

#g ( t ) = \frac{2}{3} tan t#,
#g' ( t ) = \frac{2}{3} sec^2 t#, and
#g^{-1} ( x ) = tan^{-1} \frac{3 x}{2}#.

With this, we get

# [ \frac{1}{81} int \frac{1}{(\frac{4 tan^2 t}{9} + \frac{4}{9})^2} \cdot \frac{2}{3} sec^2 t\ dt ]_{t = tan^{-1} \frac{3 x}{2}} = #

# = [ \frac{1}{24} int \frac{1}{(tan^2 t + 1)^2} \cdot sec^2 t\ dt ]_{t = tan^{-1} \frac{3 x}{2}} #.

Using the trigonometric identity mentioned before, this is the same as

# [ \frac{1}{24} int \frac{1}{sec^4 t} \cdot sec^2 t\ dt ]_{t = tan^{-1} \frac{3 x}{2}} = #

# = [ \frac{1}{24} int cos^2 t\ dt ]_{t = tan^{-1} \frac{3 x}{2}} #.

According to the double-angle formula for the cosine, #cos ( 2t ) = 2 cos^2 t - 1#, this is the same as

# [ \frac{1}{48} int cos( 2 t) + 1\ dt ]_{t = tan^{-1} \frac{3 x}{2}} #,

which is easily evaluated to be

# \frac{1}{48} [ \frac{sin(2 t)}{2} + t ]_{t = tan^{-1} \frac{3 x}{2}} + C #.

Lastly, we use the double-angle formula for the sine,

#sin (2t) = (2 tan t) / (1 + tan^2 t)#,

and thus arrive at the result

# \frac{1}{48} [ \frac{tan t}{1 + tan^2 t} + t ]_{t = tan^{-1} \frac{3 x}{2}} + C = #

# = \frac{1}{48} [ \frac{\frac{3 x}{2}}{1 + \frac{9 x^2}{4}} + tan^{-1} \frac{3 x}{2} ] + C = #

# = \frac{1}{48} [ \frac{6 x}{9 x^2 + 4} + tan^{-1} \frac{3 x}{2} ] + C #.