A) \[3+2\pi \]
B) \[4-\pi \]
C) \[2+\pi \]
D) None of these
Correct Answer: B
Solution :
Put \[{{e}^{x}}-1={{t}^{2}}\Rightarrow {{e}^{x}}dx=2t\,dt\] Also as \[x=0\]to \[\log 5,t=0\]to 2 Therefore, \[\int_{0}^{\log 5}{\frac{{{e}^{x}}\sqrt{{{e}^{x}}-1}}{{{e}^{x}}+3}}dx=\int_{0}^{2}{\frac{2{{t}^{2}}}{{{t}^{2}}+4}dt}\] \[=2\left[ \int_{0}^{2}{1dt-4\int_{0}^{2}{\frac{dt}{{{t}^{2}}+4}}} \right]=4-\pi \].
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