20.tex

\def\no{20}
\def\theintegral{\(\int{\frac{ 1 }{ u^2-a^2 } }\dif{u}%
=\frac{1}{2a} { \ln\abs{ \frac{u-a}{ u+a } } }+C\)}

% \def\aroundarrows{\mspace{18mu}}
% \def\andsoarrow{\Rightarrow}
\begin{align*}
% \renewcommand{\baselinestretch}{2}
{I}_{\no}
=&  \int{ \frac{ 1 }{ u^2-a^2 } }\dif{u} \\
=&  \int{ \frac{ A }{ u-a } + \frac{ B }{ u+a } }\dif{u}
\nb{\sub*{}}[\quad
  A(u+a)+B(u-a) = 1
  \andso A+B=0 \quad\text{and}\quad A-B=\tfrac{1}{a} \\
  \andso B=-A
  \andso 2A=\tfrac{1}{a}
  \andso A=\tfrac{1}{2a}
  \andso B=-\tfrac{1}{2a}
]
=&  \int{ \frac{ \frac{1}{2a} }{ u-a }
        + \frac{ -\frac{1}{2a} }{ u+a } }\dif{u} \\
=&  \frac{1}{2a} \cdot \int{
       \frac{ 1 }{ u-a } - \frac{ 1 }{ u+a } }\dif{u} \\
=&  \frac{1}{2a} \left(
      \ln\abs{ u-a } - \ln\abs{ u+a } + C
    \right) \\
=&  \frac{1}{2a} \cdot { \ln\abs{ \frac{u-a}{ u+a } } }+C
\end{align*}