Explanation

(b)
$\int_{}^{}{\frac{2e^{x} + 3e^{- x}}{3e^{x} + 4e^{- x}}\ dx =}\int_{}^{}{\frac{2e^{2x} + 3}{3e^{2x} + 4}dx}$
Let $2e^{2x} + 3 = A\left( 3e^{2x} + 4 \right) + B(6e^{2x})$
$\Rightarrow \ 2e^{2x} + 3 = (3A + 6B)e^{- 2x} + 4A$
On comparing both sides, we get
$2 = 3A + 6B$ …(i)
and $3 = 4A$ …(ii)
From Eqs. (i) and (ii), we get
$2 = \frac{9}{4} + 6B$
$\Rightarrow 6B = 2 - \frac{9}{4} = - \frac{1}{4}$
$\Rightarrow B = - \frac{1}{24}$
$\therefore\ \int_{}^{}{\frac{2e^{2x} + 3}{3e^{2x} + 4}dx =}\int_{}^{}{\frac{3(3e^{2x} + 4)}{4(3e^{2x} + 4)}dx - \frac{1}{24}\int_{}^{}{\frac{6e^{2x}}{3e^{2x} + 4}dx}}$
$= \frac{3}{4}x - \frac{1}{24}\log{\left( 3e^{2x} + 4 \right) + c}$