A) \[-\frac{1}{6{{x}^{3}}}\]
B) \[\frac{3}{{{x}^{2}}}\]
C) \[-\frac{1}{2{{x}^{2}}}\]
D) \[-\frac{1}{2{{x}^{3}}}\]
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{\frac{dx}{{{x}^{3}}{{\left( 1+{{x}^{6}} \right)}^{2/3}}}}=xf(x){{(1+{{x}^{6}})}^{1/3}}+c\] \[\int_{{}}^{{}}{\frac{dx}{{{x}^{7}}{{\left( \frac{1}{{{x}^{6}}}+1 \right)}^{2/3}}}}=x\,f(x){{(1+{{x}^{6}})}^{1/3}}+c\] Let\[t=\frac{1}{{{x}^{6}}}+1\] \[dt=\frac{-6}{{{x}^{7}}}dx\] \[-\frac{1}{6}\int_{{}}^{{}}{\frac{dt}{{{t}^{2/3}}}}=-\frac{1}{2}{{t}^{1/3}}\] \[=-\frac{1}{2}{{\left( \frac{1}{{{x}^{6}}}+1 \right)}^{1/3}}=-\frac{1}{2}\frac{{{\left( 1+{{x}^{6}} \right)}^{1/3}}}{{{x}^{2}}}\] \[\therefore \]\[f(x)=-\frac{1}{2{{x}^{3}}}\]
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