A) 0
B) \[\infty \]
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: B
Solution :
\[\int_{0}^{\infty }{\frac{{{x}^{3}}dx}{{{({{x}^{2}}+4)}^{2}}}=\frac{1}{2}}\int_{0}^{\infty }{\frac{{{x}^{2}}2x\,dx}{{{({{x}^{2}}+4)}^{2}}}dx}\]\[=\frac{1}{2}\int_{0}^{\infty }{\frac{t}{{{(t+4)}^{2}}}dt}\], [Putting \[{{x}^{2}}=t\]] \[=\frac{1}{2}\int_{0}^{\infty }{\left[ \frac{1}{t+4}-\frac{4}{{{(t+4)}^{2}}} \right]dt=\frac{1}{2}\left[ \log (t+4)+\frac{4}{t+4} \right]_{0}^{\infty }}\] \[=\frac{1}{2}\left[ \log \infty +0-(\log 4+1) \right]=\infty \].
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