A) \[\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}+c\]
B) \[\log ({{e}^{2x}}+1)-x+c\]
C) \[\log ({{e}^{2x}}+1)+c\]
D) None of these
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}\,dx}=\int_{{}}^{{}}{\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}\,dx}\] 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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