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

Question

William Abdalla · Accepted Answer

Quick answer :

#int1/(x^2-4)^2dx = 1/16int1/(-1/4x^2+1)dx#

substitute #u = 1/2x#

#du = 1/2#

#u^2 = 1/4x^2#

So #1/8int 1/(-u^2+1)^2du#

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

let's #u = tanh(t)#

#du = 1/cosh^2(t)dt#

#-u^2 = -tanh^2(t)#

don't forget #1-tanh^2(t) = 1/cosh^2(t)#

so we have

#1/8int1/(1/cosh^2(t))^2*1/cosh^2(t)dt = 1/8intcosh^2(t)#

dont forget #cosh^2(t) = 1/2(1+cosh(2t))#

#1/16int1+cosh(2t)dt#

#1/16[t+1/2sinh(2t)]+C#

and then substitute back

Emilia Aguilar · Answer

undefined To find the integral of $ \frac{dx}{(x^2 - 4)^2} $, you can use partial fraction decomposition followed by integration techniques.

1. Decompose the fraction into partial fractions:
$$ \frac{1}{(x^2 - 4)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{x-2} + \frac{D}{(x-2)^2} $$

2. Solve for the constants $ A $, $ B $, $ C $, and $ D $.

3. Once you have the partial fraction decomposition, integrate each term separately.

4. The integral of $ \frac{A}{x+2} $ would be $ A\ln|x+2| $, the integral of $ \frac{B}{(x+2)^2} $ would be $ -\frac{B}{x+2} $, the integral of $ \frac{C}{x-2} $ would be $ C\ln|x-2| $, and the integral of $ \frac{D}{(x-2)^2} $ would be $ -\frac{D}{x-2} $.

5. Finally, sum up all the integrals to get the solution.

The integral of $ \frac{dx}{(x^2 - 4)^2} $ involves applying these steps to find the integral in terms of logarithmic and inverse functions.