# How do you find the integral of  dx / (x^2 - 4)^2?

Jul 1, 2015

$\int \frac{1}{{x}^{2} - 4} ^ 2 \mathrm{dx} = \frac{1}{16} \int \frac{1}{- \frac{1}{4} {x}^{2} + 1} \mathrm{dx}$

substitute $u = \frac{1}{2} x$

$\mathrm{du} = \frac{1}{2}$

${u}^{2} = \frac{1}{4} {x}^{2}$

So $\frac{1}{8} \int \frac{1}{- {u}^{2} + 1} ^ 2 \mathrm{du}$

Here you can do partial fraction but it's long...

let's $u = \tanh \left(t\right)$

$\mathrm{du} = \frac{1}{\cosh} ^ 2 \left(t\right) \mathrm{dt}$

$- {u}^{2} = - {\tanh}^{2} \left(t\right)$

don't forget $1 - {\tanh}^{2} \left(t\right) = \frac{1}{\cosh} ^ 2 \left(t\right)$

so we have

$\frac{1}{8} \int \frac{1}{\frac{1}{\cosh} ^ 2 \left(t\right)} ^ 2 \cdot \frac{1}{\cosh} ^ 2 \left(t\right) \mathrm{dt} = \frac{1}{8} \int {\cosh}^{2} \left(t\right)$

dont forget ${\cosh}^{2} \left(t\right) = \frac{1}{2} \left(1 + \cosh \left(2 t\right)\right)$

$\frac{1}{16} \int 1 + \cosh \left(2 t\right) \mathrm{dt}$

$\frac{1}{16} \left[t + \frac{1}{2} \sinh \left(2 t\right)\right] + C$

and then substitute back