If \[\int_{0}^{k}{\frac{dx}{2+8{{x}^{2}}}}=\frac{\pi }{16}\,,\] then \[k=\]

If \[\int_{0}^{k}{\frac{dx}{2+8{{x}^{2}}}}=\frac{\pi }{16}\,,\] then \

1.	If \[\int_{0}^{k}{\frac{dx}{2+8{{x}^{2}}}}=\frac{\pi }{16}\,,\] then \[k=\]
A.	1
B.	\[\frac{1}{2}\]
C.	\[\frac{1}{4}\]
D.	None of these
Answer» C. \[\frac{1}{4}\]

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If \[\int_{0}^{k}{\frac{dx}{2+8{{x}^{2}}}}=\frac{\pi }{16}\,,\] then \[k=\]

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