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

2 Answers
Apr 19, 2018

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

Explanation:

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.

Apr 19, 2018

int \frac{1}{(9 x^2 + 4)^2} dx = \frac{1}{48} [\frac{6 x}{9 x^2 + 4} + tan^{-1} \frac{3 x}{2} ] + C

Explanation:

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 .