Question

Find the integrals of the following:

${\displaystyle \frac{1}{4-x^2}}$

Answer 1

${\displaystyle \int \frac{1}{4-x^2} \text{d}x=\int \frac{1}{{\left(2\right)}^2-{\left(x\right)}^2} \text{d}x}$

${\displaystyle \int \frac{1}{{\text{a}}^2-x^2} \text{d}x=\frac{1}{2\text{a}}\log \left|\frac{\text{a}+x}{\text{a}-x}\right|+\text{c}}$

${\displaystyle \int \frac{1}{4-x^2} \text{d}x=\frac{1}{2\left(2\right)}\log \left|\frac{2+x}{2-x}\right|+\text{c}}$

= ${\displaystyle \frac{1}{4}\log \left|\frac{2+x}{2-x}\right|+\text{c}}$