A) \[\frac{1}{3}{{({{x}^{2}}+2)}^{3/2}}+2{{({{x}^{2}}+2)}^{1/2}}+c\]
B) \[\frac{1}{3}{{({{x}^{2}}+2)}^{3/2}}-2{{({{x}^{2}}+2)}^{1/2}}+c\]
C) \[\frac{1}{3}{{({{x}^{2}}+2)}^{3/2}}+{{({{x}^{2}}+2)}^{1/2}}+c\]
D) \[\frac{1}{3}{{({{x}^{2}}+2)}^{3/2}}-{{({{x}^{2}}+2)}^{1/2}}+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{{{x}^{3}}}{\sqrt{{{x}^{2}}+2}}\,dx}=\int_{{}}^{{}}{\frac{{{x}^{2}}.\,x}{\sqrt{{{x}^{2}}+2}}\,dx}\] Put \[{{x}^{2}}+2={{t}^{2}}\Rightarrow x\,dx=t\,dt\] and \[{{x}^{2}}={{t}^{2}}-2,\] then it reduces to \[\int_{{}}^{{}}{\frac{({{t}^{2}}-2)}{t}\,t\,dt}=\int_{{}}^{{}}{({{t}^{2}}-2)dt}\] \[=\frac{{{t}^{3}}}{3}-2t+c=\frac{{{({{x}^{2}}+2)}^{3/2}}}{3}-2{{({{x}^{2}}+2)}^{1/2}}+c.\]
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