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

MCQ

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

✓
$\log ({e^x} - {e^{ - x}}) + c$
B
$\log ({e^x} + {e^{ - x}}) + c$
C
$\log ({e^{ - x}} - {e^x}) + c$
D
$\log (1 - {e^{ - x}}) + c$

✓

Answer

Correct option: A.

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

a
(a) $I = \int_{}^{} {\frac{{{e^{2x}} + 1}}{{{e^{2x}} - 1}}} = \int_{}^{} {\frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}} \,dx$
Put ${e^x} - {e^{ - x}} = t \Rightarrow ({e^x} + {e^{ - x}})\,dx = dt$
$\therefore \,\,\,I = \int_{}^{} {\frac{{dt}}{t}\,dt} = \log t + c = \log ({e^x} - {e^{ - x}}) + c$.

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