Q.Prove that: $\displaystyle\int \dfrac{1}{a^2-x^2}\,dx=\dfrac{1}{2a}\log\left(\dfrac{a+x}{a-x}\right)+c$.

A general quadratic denominator (or quadratic under a square root) is reduced to one of the seven special integrals by a five-step routine: pull the leading coefficient $a$ (or $\sqrt a$) outside; complete the square on the remaining $x^2+\frac bax$ by adding and subtracting $\left(\frac12\cdot\text{coefficient of }x\right)^2$; write the result as a sum or difference of two squares, $(x+\beta)^2\pm\alpha^2$; shift the standard formula (since $\int f(x)\,dx=g(x)+c\Rightarrow\int f(x+\beta)\,dx=g(x+\beta)+c$); and finally apply whichever of the seven special formulae matches. Whether the final answer is an inverse tangent, a logarithm, or an inverse sine depends e …