_ ^ e^ 2x - 1 e^ 2x + 1 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} \;dx = $

A
$\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}} + c$
✓
$\log ({e^{2x}} + 1) - x + c$
C
$\log ({e^{2x}} + 1) + c$
D
None of these

✓

Answer

Correct option: B.

$\log ({e^{2x}} + 1) - x + c$

b
(b)$\int_{}^{} {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}\,dx} = \int_{}^{} {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}\,dx} $
Now put ${e^x} + {e^{ - x}} = t \Rightarrow ({e^x} - {e^{ - x}})dx = dt,$ then it reduces to $\int_{}^{} {\frac{{dt}}{t}} = \log t = \log ({e^x} + {e^{ - x}}) = \log ({e^{2x}} + 1) - x + c$.

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