7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

Put $1 + x = {t^2} \Rightarrow dx = 2t\,dt$

When $x = 3 \to 8,$ then $t = 2 \to 3$

$\therefore$  $I = 2\int_2^3 {\frac{{5 - 3{t^2}}}{{{t^2} - 1}}dt} $;

$I = 2\int_2^3 {\left( {\frac{2}{{{t^2} - 1}} - 3} \right)} \,dt$

$I = 2\left[ {\frac{2}{{2.1}}\log \frac{{t - 1}}{{t + 1}} - 3t} \right]_2^3$;

$I = 2\log \left( {\frac{3}{{2{e^3}}}} \right)$.