Question

Good afternoon, please how can I integrate this function? dx/(4x^2-9)^2

Answer 1

HI!!

Answer 2

square it and you will get an easy polynomial to integrate

Answer 3

can you explain more please? I thought that I can solve by trig. sustitu

Answer 4

You can proceed with either a trigo sub or partial fractions. Let's try the former. Set \(x=\dfrac{3}{2}\sec t\), then \(dx=\dfrac{3}{2}\sec t\tan t\,dt\).
\[\begin{align*}
I&=\int\frac{dx}{(4x^2-9)^2}\\\\

&=\frac{3}{2}{\Huge\int}\frac{\sec t\tan t}{\left(4\left(\frac{3}{2}\sec t\right)^2-9\right)^2}\,dt\\\\

&=\frac{3}{2}{\Large\int}\frac{\sec t\tan t}{\left(4\times\frac{9}{4}\sec^2 t-9\right)^2}\,dt\\\\

&=\frac{3}{2}{\Large\int}\frac{\sec t\tan t}{9^2\left(\sec^2 t-1\right)^2}\,dt\\\\

&=\frac{1}{54}{\Large\int}\frac{\sec t\tan t}{\left(\tan^2 t\right)^2}\,dt\\\\

&=\frac{1}{54}\int\frac{\sec t}{\tan^3 t}\,dt\\\\

&=\frac{1}{54}\int\frac{\cos^2t}{\sin^3t}\,dt\\\\

&=\frac{1}{54}\int\frac{1-\sin^2t}{\sin^3t}\,dt\\\\

&=\frac{1}{54}\left(\int\csc^3t\,dt-\int\csc t\,dt\right)
\end{align*}\]
You can integrate by parts to derive the reduction formula for the integral of \(\csc^nt\) below:
\[\int\csc^n t\,dt=-\frac{\csc^{n-2}t\cot t}{n-1}+\frac{n-2}{n-1}\int\csc^{n-2}t\,dt\]
This gives
\[\begin{align*}
I&=\frac{1}{54}\left(-\frac{\csc t\cot t}{2}+\frac{1}{2}\int\csc t\,dt-\int\csc t\,dt\right)\\\\

&=-\frac{1}{108}\left(\csc t\cot t+\int\csc t\,dt\right)
\end{align*}\]

Answer 5

The latter approach, using partial fractions:
\[\begin{align*}
\frac{1}{(4x^2-9)^2}&=\frac{1}{(2x-3)^2(2x+3)^2}\\\\

&=\frac{A}{2x-3}+\frac{B}{(2x-3)^2}+\frac{C}{2x+3}+\frac{D}{(2x+3)^2}\\\\

1&=A(2x-3)(2x+3)^2+B(2x+3)^2\\
&\quad\quad+C(2x-3)^2(2x+3)+D(2x-3)^2\\\\

&=A(8 x^3+12 x^2-18 x-27)+B(4 x^2+12 x+9)\\
&\quad\quad+C(8 x^3-12 x^2-18 x+27)+D(4 x^2-12 x+9)\\\\

&=(8A+8C)x^3+(12A+4B-12C+4D)x^2\\
&\quad\quad+(-18A+12B-18C-12D)x+(-27A+9B+27C+9D)
\end{align*}\]
Matching coefficients yields the system below, with solution:
\[\begin{cases}8A+8C=0\\
12A+4B-12C+4D=0\\
-18A+12B-18C-12D=0\\
-27A+9B+27C+9D=1\end{cases}~~\implies~~\begin{matrix}A=-\dfrac{1}{108}&&&
B=\dfrac{1}{36}\\
C=\dfrac{1}{108}&&&
D=\dfrac{1}{36}\end{matrix}\]
To summarize,
\[\begin{align*}\frac{1}{(4x^2-9)^2}&=-\frac{1}{108}\left(\frac{1}{2x-3}-\frac{3}{(2x-3)^2}-\frac{1}{2x+3}-\frac{3}{(2x+3)^2}\right)\\\\
\int\frac{1}{(4x^2-9)^2}\,dx&=-\frac{1}{108}\left(\int+\cdots+\int\right)\end{align*}\]