Question

what is the integral of 1/(16x^2-1)

Answer 1

Let \(x=\dfrac{1}{4}\sec u\), so that \(dx=\dfrac{1}{4}\sec u\tan u~du\).
\[\int\frac{1}{16x^2-1}~dx=\int\frac{\dfrac{1}{4}\sec u\tan u}{16\left(\dfrac{1}{4}\sec u\right)^2-1}~du=\frac{1}{4}\int\frac{\sec u\tan u}{\sec^2 u-1}~du\]

Answer 2

another approach is to split it into 2 fractions
\[\frac{1}{16x^{2} -1} = \frac{1}{(4x+1)(4x-1)} = \frac{A}{4x+1} + \frac{B}{4x-1}\]
where
\[A(4x-1) + B(4x+1) = 1\]
4A +4B = 0 --> A = -B
B -A = 1 --> B = 1/2
--> A = -1/2
integral becomes
\[\frac{1}{2} \int\limits \frac{1}{4x-1} - \frac{1}{4x+1} dx\]
use \[\int\limits \frac{1}{u} = \ln u\]