A) \[x-\log [1+\sqrt{1-{{e}^{2x}}}]+c\]
B) \[x+\log [1+\sqrt{1-{{e}^{2x}}}]+c\]
C) \[\log [1+\sqrt{1-{{e}^{2x}}}]-x+c\]
D) None of these
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\frac{1}{\sqrt{1-{{e}^{2x}}}}\,dx=\int_{{}}^{{}}{\frac{{{e}^{-x}}}{\sqrt{{{e}^{-2x}}-1}}\,dx}}\] Put \[{{e}^{-x}}=t\Rightarrow -{{e}^{-x}}dx=dt,\] then it reduces to \[-\int_{{}}^{{}}{\frac{1}{\sqrt{{{t}^{2}}-1}}\,dt=-\log \left[ t+\sqrt{{{t}^{2}}-1} \right]+c}\] \[=-\log \left[ {{e}^{-x}}+\sqrt{{{e}^{-2x}}-1} \right]=-\log \left[ \frac{1}{{{e}^{x}}}+\frac{\sqrt{1-{{e}^{2x}}}}{{{e}^{x}}} \right]\] \[=-\log \left[ 1+\sqrt{1-{{e}^{2x}}} \right]+\log {{e}^{x}}+c\] \[=x-\log \left[ 1+\sqrt{1-{{e}^{2x}}} \right]+c.\]
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