A) 1
B) 2
C) 4
D) -1
Correct Answer: C
Solution :
\[\int_{\log 2}^{x}{\frac{du}{{{({{e}^{u}}-1)}^{1/2}}}}=\frac{\pi }{6}\] Þ \[\int_{1}^{\sqrt{{{e}^{x}}-1}}{\frac{2t}{1+{{t}^{2}}}}\ dt=\frac{\pi }{6}\]as \[{{e}^{u}}-1={{t}^{2}}\] Þ \[2({{\tan }^{-1}}t)_{1}^{\sqrt{{{e}^{x}}-1}}=\frac{\pi }{6}\] Þ \[{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}-\frac{\pi }{4}=\frac{\pi }{12}\] Þ \[\sqrt{{{e}^{x}}-1}=\tan \frac{\pi }{3}\] Þ \[\sqrt{{{e}^{x}}-1}=\sqrt{3}\] Þ \[{{e}^{x}}=4\].
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