Derive these formulas: $$\bbox[5px, border:1px solid black, #F9F7ED]{{(a)} \quad \int \dfrac{dx}{x^{2}-a^{2}}=\dfrac{1}{2a}\ln \left \vert \dfrac{x-a}{x+a}\right \vert +C \quad a ≠ 0 }$$ $$\bbox[5px, border:1px solid black, #F9F7ED]{{(b)} \quad \int \dfrac{dx}{a^{2}-x^{2}}=\dfrac{1}{2a}\ln \left\vert \dfrac{x+a}{x-a}\right\vert +C \quad a ≠ 0 }$$

Solution  (a) The factored denominator $x^{2}-a^{2}=( x-a) (x+a)$ contains only distinct linear factors. So, $\dfrac{1}{x^{2}-a^{2}}$ can be decomposed into partial fractions of the form $$\begin{eqnarray*} \begin{array}{rl@{\qquad}l} \dfrac{1}{x^{2}-a^{2}} &= \dfrac{1}{(x-a)(x+a)}=\dfrac{A}{x-a}+\dfrac{B}{x+a}\\ 1 &= A(x+a) +B (x-a)&{\color{#0066A7}{\hbox{Multiply both sides by }(x-a) (x+a).}} \end{array} \end{eqnarray*}$$

This is an identity in $x.$ When $x=a,$ the term involving $B$ drops out. Then $1=A(2a)$, so $A=\dfrac{1}{2a}.$ When $x=-a$, the term involving $A$ drops out. Then $1=B (-2a)$, so $B=-\dfrac{1}{2a}.$ Then, $$\begin{equation*} \dfrac{1}{x^{2}-a^{2}}=\dfrac{A}{x-a}+\dfrac{B}{x+a}=\dfrac{1}{2a( x-a) }-\dfrac{1}{2a(x+a) } \end{equation*}$$ $$\begin{eqnarray*} \int \dfrac{dx}{x^{2}-a^{2}} &=& \dfrac{1}{2a}\int \dfrac{dx}{x-a}-\dfrac{1}{2a} \int \dfrac{dx}{x+a}=\dfrac{1}{2a}\left( \int \dfrac{dx}{x-a}-\int \dfrac{dx}{x+a}\right) \\[7pt] &=& \dfrac{1}{2a}\big(\ln \,\vert x-a\vert -\ln \,\vert x+a \vert\big) +C\\[7pt] &=&\dfrac{1}{2a}\ln \left\vert \dfrac{x-a}{x+a}\right\vert +C \end{eqnarray*}$$

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(b) Using the result from (a), we get $$\begin{eqnarray*} \int \dfrac{dx}{a^{2}-x^{2}} &=& -\int \dfrac{dx}{x^{2}-a^{2}}=-\dfrac{1}{2a}\ln \left\vert \dfrac{x-a}{x+a}\right\vert +C=\dfrac{1}{2a}\ln \left\vert \dfrac{x-a }{x+a}\right\vert ^{-1}+C \\[6pt] &=& \dfrac{1}{2a}\ln \left\vert \dfrac{x+a}{x-a}\right\vert +C \end{eqnarray*}$$