Scale difference-of-squares logarithmic forms

Write $4=2^2$. Then $\int\frac1{x^2-4}\,dx=\frac1{2(2)}\ln\left|\frac{x-2}{x+2}\right|+C=\frac14\ln\left|\frac{x-2}{x+2}\right|+C$, for $x\ne\pm2$.

$\int\frac1{4x^2-1}\,dx=\frac14\int\frac1{x^2-(1/2)^2}\,dx=\frac14\frac1{2(1/2)}\ln\left|\frac{x-1/2}{x+1/2}\right|+C=\frac14\ln\left|\frac{2x-1}{2x+1}\right|+C$, for $x\ne\pm1/2$.

The handwritten alternative takes $f(x)=2x$ and $f^{\prime}(x)=2$: $\int\frac1{(2x)^2-1}\,dx=\frac12\int\frac2{(2x)^2-1}\,dx=\frac12\frac12\ln\left|\frac{2x-1}{2x+1}\right|+C$. This gives the same quarter-logarithm.

$\int\frac1{16-4x^2}\,dx=\frac14\int\frac1{2^2-x^2}\,dx=\frac14\frac1{2(2)}\ln\left|\frac{2+x}{2-x}\right|+C=\frac1{16}\ln\left|\frac{2+x}{2-x}\right|+C$, for $x\ne\pm2$.