A) \[A=\frac{1}{4},B=\frac{1}{5}\]
B) \[A=\frac{1}{8},B=-\frac{1}{5}\]
C) \[A=-\frac{1}{8},B=\frac{1}{5}\]
D) none of these
Correct Answer: B
Solution :
[b] Here, \[\int{{{x}^{5}}{{(1+{{x}^{3}})}^{2/3}}dx}\] Let \[1+{{x}^{3}}={{t}^{2}}\]and \[3{{x}^{2}}dx=2tdt\] \[\therefore \int{{{x}^{5}}{{(1+{{x}^{3}})}^{2/3}}dx=\int{{{x}^{3}}{{(1+{{x}^{3}})}^{2/3}}{{x}^{2}}dx}}\] \[=\int{({{t}^{2}}-1){{({{t}^{2}})}^{2/3}}{{x}^{2}}dx}\] \[=\frac{2}{3}\int{({{t}^{2}}-1){{t}^{7/3}}dt}\] \[=\frac{2}{3}\int{({{t}^{13/3}}-{{t}^{7/3}})\,dt}\] \[=\frac{2}{3}\left\{ \frac{3}{16}{{t}^{16/3}}-\frac{3}{10}{{t}^{10/3}} \right\}+C\] \[=\frac{1}{8}{{(1+{{x}^{3}})}^{8/3}}-\frac{1}{5}{{(1+{{x}^{3}})}^{5/3}}+C\]
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