A) \[a\,(\sqrt{2}-1)\]
B) \[a\,(1-\sqrt{2})\]
C) \[a\,(1+\sqrt{2})\]
D) \[2a\sqrt{3}\]
Correct Answer: A
Solution :
Put \[t={{a}^{2}}+{{x}^{2}}\Rightarrow 2xdx=dt,\]then \[\int_{0}^{a}{\frac{xdx}{\sqrt{{{a}^{2}}+{{x}^{2}}}}=\frac{1}{2}\int_{{{a}^{2}}}^{2{{a}^{2}}}{\frac{1}{\sqrt{t}}dt}}\] \[=[{{(2{{a}^{2}})}^{1/2}}-{{a}^{2/2}}]=a(\sqrt{2}-1)\].
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