$\cfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}$
$\cfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}$
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NCERT & Exemplar
: Let $I = \int {\cfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}dx} = \int {\cfrac{{{e^x}\left( {{e^x} - {e^{ - x}}} \right)}}{{{e^x}\left( {{e^x} + {e^{ - x}}} \right)}}dx} = \int {\cfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} dx$
Put ${e^x} + {e^{ - x}} = t$ $\Rightarrow$ $\left( {{e^x} - {e^{ - x}}} \right)dx = dt$
$\therefore$ $I = \int {\cfrac{{dt}}{t} = \log \left| t \right| + C} = \log \left| {{e^x} + {e^{ - x}}} \right| + C$
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