A) \[2{{e}^{4}}-2e-a\]
B) \[2{{e}^{4}}-e-a\]
C) \[2{{e}^{4}}-e-2a\]
D) \[{{e}^{4}}-e-a\]
Correct Answer: B
Solution :
[b] \[{{I}_{1}}=\int_{e}^{{{e}^{4}}}{\sqrt{\ln x}dx}\] Putting \[t=\sqrt{\ln x},\]i.e., \[dt=\frac{dx}{2x\sqrt{\ln x}},\] we get \[dx=2t\,{{e}^{{{t}^{2}}}}dt\] Or \[\int_{e}^{{{e}^{4}}}{\sqrt{\ln x}\,}dx=\int_{1}^{2}{2{{t}^{2}}{{e}^{{{t}^{2}}}}dt}\] \[=t\,{{e}^{{{t}^{2}}}}|_{1}^{2}-\int_{1}^{2}{{{e}^{{{t}^{2}}}}dt}\] \[=2{{e}^{4}}-e-a\]
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