A) \[\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}\,-\frac{{{a}^{2}}}{2}\log \{x+\sqrt{{{x}^{2}}+{{a}^{2}}}\}+c\]
B) \[\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}\,+\frac{{{a}^{2}}}{2}\log \,\{x+\sqrt{{{x}^{2}}+{{a}^{2}}}\}+c\]
C) \[\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}\,-\frac{{{a}^{2}}}{2}\log \{x-\sqrt{{{x}^{2}}+{{a}^{2}}}\}+c\]
D) \[\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}\,+\frac{{{a}^{2}}}{2}\log \{x-\sqrt{{{x}^{2}}+{{a}^{2}}}\}+c\]
Correct Answer: B
Solution :
\[I=\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,dx}\]\[=\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,.1\,dx}\] \[=\sqrt{{{x}^{2}}+{{a}^{2}}}\int{1dx-\int{\left[ \frac{d}{dx}\left( \sqrt{{{x}^{2}}+{{a}^{2}}} \right)\int{1\,dx} \right]\,dx}}\] \[=x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \frac{2x}{2\sqrt{{{x}^{2}}+{{a}^{2}}}}x \right]}\,dx\] \[=x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \frac{{{x}^{2}}+{{a}^{2}}-{{a}^{2}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]}dx\] \[=x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \sqrt{{{x}^{2}}+{{a}^{2}}}-\frac{{{a}^{2}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]\,}dx\] \[=x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx+{{a}^{2}}\int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}}\] \[=\sqrt{{{x}^{2}}+{{a}^{2}}}-I+{{a}^{2}}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C\] Þ \[2I=x\sqrt{{{x}^{2}}+{{a}^{2}}+}{{a}^{2}}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C\] Þ \[I=\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{{{a}^{2}}}{2}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C\].
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