A) \[\frac{1}{{{a}^{4}}}\left[ \frac{1}{x}\sqrt{{{a}^{2}}+{{x}^{2}}}-\frac{1}{3{{x}^{2}}}\sqrt{{{a}^{2}}+{{x}^{2}}} \right]+c\]
B) \[\frac{1}{{{a}^{4}}}\left[ \frac{1}{x}\sqrt{{{a}^{2}}+{{x}^{2}}}-\frac{1}{3{{x}^{3}}}{{({{a}^{2}}+{{x}^{2}})}^{3/2}} \right]+c\]
C) \[\frac{1}{{{a}^{4}}}\left[ \frac{1}{x}\sqrt{{{a}^{2}}+{{x}^{2}}}-\frac{1}{2\sqrt{x}}{{({{a}^{2}}+{{x}^{2}})}^{3/2}} \right]+c\]
D) \[\frac{1}{{{a}^{4}}}\left[ \frac{1}{x}\sqrt{{{a}^{2}}+{{x}^{2}}}+\frac{1}{3{{x}^{3}}}\sqrt{{{a}^{2}}+{{x}^{2}}} \right]+c\]
Correct Answer: B
Solution :
\[{{x}^{4}}\sqrt{{{a}^{2}}+{{x}^{2}}}={{x}^{5}}\sqrt{{{\left( \frac{a}{x} \right)}^{2}}+1}\] Now, \[I=\int_{{}}^{{}}{\frac{\frac{dx}{{{x}^{3}}}}{{{x}^{2}}\sqrt{{{\left( \frac{a}{x} \right)}^{2}}+1}}}\] Let \[\frac{{{a}^{2}}}{{{x}^{2}}}=t,\]so \[\frac{-2{{a}^{2}}}{{{x}^{3}}}dx=dt\] \[I=-\frac{1}{2{{a}^{4}}}\int_{{}}^{{}}{\frac{t}{\sqrt{t+1}}}dt=-\frac{1}{2{{a}^{4}}}\int_{{}}^{{}}{\frac{t+1-1}{\sqrt{t+1}}dt}\] \[=\frac{1}{{{a}^{4}}}\left[ \sqrt{t+1}-\frac{1}{3}{{(t+1)}^{3/2}} \right]+c\] \[=\frac{1}{{{a}^{4}}}\left[ \frac{\sqrt{{{a}^{2}}+{{x}^{2}}}}{x}-\frac{1}{3{{x}^{3}}}{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{3/2}} \right]+c\]
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