Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{{{e^{2x}} - {e^{ - 2x}}}}{{{e^{2x}} + {e^{ - 2x}}}}dx}$

Put ${e^{2x}} + {e^{ - 2x}} = t$
$\Rightarrow$ $\left( {2{e^{2x}} - 2{e^{ - 2x}}} \right)dx = dt$ $\Rightarrow$ $\left( {{e^{2x}} - {e^{ - 2x}}} \right)dx = \cfrac{{dt}}{2}$

$\therefore$ $I = \cfrac{1}{2}\int {\cfrac{1}{t}dt} = \cfrac{1}{2}\log \left| t \right| + C = \cfrac{1}{2}\log \left| {{e^{2x}} + {e^{ - 2x}}} \right| + C$