\[\boxed{\text{199\ (199).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\frac{4x + 3}{x^{2} - 1} = \frac{a^{\backslash x + 1}}{x - 1} + \frac{b^{\backslash x - 1}}{x + 1}\]
\[Преобразуем\ правую\ часть\ \]
\[равенства:\]
\[\frac{4x + 3}{x^{2} - 1} = \frac{ax + a + bx - b}{(x - 1)(x + 1)}\]
\[4x + 3 = x(a + b) + (a - b)\]
\[\left\{ \begin{matrix} a + b = 4 \\ a - b = 3 \\ \end{matrix} \right.\ \]
\[a + b + a - b = 4 + 3\]
\[2a = 7\]
\[a = 3,5.\]
\[a + b = 4\]
\[3,5 + b = 4\]
\[b = 0,5.\]
\[То\ есть\ получаем\ тождество:\ \]
\[\frac{4x + 3}{x^{2} - 1} = \frac{3,5}{x - 1} + \frac{0,5}{x + 1}.\]