${\displaystyle \int \frac{2x^2+3}{(x^2–1)(x^2–4)}dx=\log {\left(\frac{x–2}{x+2}\right)}^a{\left(\frac{x+1}{x–1}\right)}^b}+c$, then $a,b$ are respectively :

    =$\int \frac{2x^2+3}{(x^2–1)(x^2–4)}dx$

  $=\frac{11}{3}\int \frac{dx}{x^2–4}–\frac{5}{3}\int \frac{dx}{x^2–1}$

  $=\frac{11}{12}\log \left(\frac{x–2}{x+2}\right)–\frac{5}{6}\log \left(\frac{x-1}{x+1}\right)+c$

    $=\log {\left(\frac{x–2}{x+2}\right)}^{11/2}{\left(\frac{x+1}{x–1}\right)}^{5/6}+c$

$\therefore a=11/12$ and $b=5/6$.