Question
$\int_{}^{} {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} \;dx = $
${e^x} + {e^{ - x}} = t $रखने पर $ ({e^x} - {e^{ - x}})dx = dt,$
$\int_{}^{} {\frac{{dt}}{t}} = \log t = \log ({e^x} + {e^{ - x}}) = \log ({e^{2x}} + 1) - x + c$.
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