A) \[\frac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}+C\]
B) \[\log \frac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}+C\]
C) \[\log |{{e}^{x}}+{{e}^{-x}}|+C\]
D) \[\log |{{e}^{x}}-{{e}^{-x}}|+C\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}}dx=\int_{{}}^{{}}{\frac{({{e}^{x}}.{{e}^{x}}-1)}{({{e}^{x}}.{{e}^{x}}+1)}}dx\] \[=\int_{{}}^{{}}{\frac{{{e}^{x}}\left( {{e}^{x}}-\frac{1}{{{e}^{x}}} \right)}{{{e}^{x}}\left( {{e}^{x}}+\frac{1}{{{e}^{x}}} \right)}.dx}\] \[=\int_{{}}^{{}}{\frac{{{e}^{x}}-{{e}^{x}}}{{{e}^{x}}+{{e}^{-x}}}}.dx\] Let \[({{e}^{x}}+{{e}^{-x}})=t\] \[\Rightarrow \] \[{{e}^{x}}-{{e}^{-x}}=\frac{dt}{dx}\] \[\Rightarrow \] \[dx=\frac{dt}{{{e}^{x}}-{{e}^{-x}}}\] \[\int_{{}}^{{}}{\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}}dx=\int_{{}}^{{}}{\frac{{{e}^{x}}-{{e}^{-x}}}{t}}.\frac{dt}{{{e}^{x}}-{{e}^{-x}}}\] \[=\int_{{}}^{{}}{\frac{dt}{t}}\] \[\log |t|+C\] \[=\log |{{e}^{x}}+{{e}^{-x}}|+C\]
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