A) \[\log ({{e}^{x}}-{{e}^{-x}})+c\]
B) \[\log ({{e}^{x}}+{{e}^{-x}})+c\]
C) \[\log ({{e}^{-x}}-{{e}^{x}})+c\]
D) \[\log (1-{{e}^{-x}})+c\]
Correct Answer: A
Solution :
\[I=\int_{{}}^{{}}{\frac{{{e}^{2x}}+1}{{{e}^{2x}}-1}}=\int_{{}}^{{}}{\frac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}}\,dx\] Put \[{{e}^{x}}-{{e}^{-x}}=t\Rightarrow ({{e}^{x}}+{{e}^{-x}})\,dx=dt\] \[\therefore \,\,\,I=\int_{{}}^{{}}{\frac{dt}{t}\,dt}=\log t+c=\log ({{e}^{x}}-{{e}^{-x}})+c\].You need to login to perform this action.
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