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

Apr 19, 2018

$\frac{1}{48} \left\{a r c \tan \left(\frac{3 x}{2}\right) + \frac{6 x}{9 {x}^{2} + 4}\right\} + C$.

#### Explanation:

Let, $I = \int \frac{1}{9 {x}^{2} + 4} ^ 2 \mathrm{dx}$.

$9 {x}^{2} + 4 = {\left(3 x\right)}^{2} + {2}^{2}$, suggests that, $3 x = 2 \tan y$ substitution

may work.

So, let, $3 x = 2 \tan y . \therefore 3 \mathrm{dx} = 2 {\sec}^{2} y \mathrm{dy}$.

$\therefore I = \int \frac{1}{4 {\tan}^{2} y + 4} ^ 2 \cdot \frac{2}{3} {\sec}^{2} y \mathrm{dy}$,

$= \frac{2}{48} \int {\sec}^{2} \frac{y}{\sec} ^ 4 y \mathrm{dy}$,

$= \frac{1}{24} \int {\cos}^{2} y \mathrm{dy}$,

$= \frac{1}{24} \int \frac{1 + \cos 2 y}{2} \mathrm{dy}$,

$= \frac{1}{48} \left\{y + \frac{\sin 2 y}{2}\right\}$,

$= \frac{1}{48} \left\{y + \frac{1}{2} \cdot \frac{2 \tan y}{1 + {\tan}^{2} y}\right\}$,

Here, $3 x = 2 \tan y \Rightarrow y = a r c \tan \left(\frac{3 x}{2}\right)$.

$\therefore I = \frac{1}{48} \left\{a r c \tan \left(\frac{3 x}{2}\right) + \frac{\frac{3 x}{2}}{1 + {\left(\frac{3 x}{2}\right)}^{2}}\right\}$.

$\Rightarrow I = \frac{1}{48} \left\{a r c \tan \left(\frac{3 x}{2}\right) + \frac{6 x}{9 {x}^{2} + 4}\right\} + C$.

Apr 19, 2018

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

#### 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 $\left(\tan x\right) ' = {\sec}^{2} x$. Therefore, we make the substitution $x \setminus \mapsto g \left(t\right)$ with

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

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, $\cos \left(2 t\right) = 2 {\cos}^{2} t - 1$, 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,

$\sin \left(2 t\right) = \frac{2 \tan t}{1 + {\tan}^{2} t}$,

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 .