Deriving Formulas Involving a Rational Function

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, \[ \begin{equation*} \dfrac{1}{x^{2}-a^{2}}=\dfrac{A}{x-a}+\dfrac{B}{x+a}=\dfrac{1}{2a( x-a) }-\dfrac{1}{2a(x+a) } \end{equation*} \] \[ \begin{eqnarray*} \int \dfrac{dx}{x^{2}-a^{2}} &=& \dfrac{1}{2a}\int \dfrac{dx}{x-a}-\dfrac{1}{2a} \int \dfrac{dx}{x+a}=\dfrac{1}{2a}\left( \int \dfrac{dx}{x-a}-\int \dfrac{dx}{x+a}\right) \\[7pt] &=& \dfrac{1}{2a}\big(\ln \,\vert x-a\vert -\ln \,\vert x+a \vert\big) +C\\[7pt] &=&\dfrac{1}{2a}\ln \left\vert \dfrac{x-a}{x+a}\right\vert +C \end{eqnarray*} \]

502

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