A) \[\frac{{{x}^{5m}}-{{x}^{4m}}}{2m{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C\]
B) \[\frac{2m({{x}^{5m}}+{{x}^{4m}})}{{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C\]
C) \[\frac{{{x}^{4m}}}{2m{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C\]
D) \[\frac{{{x}^{5m}}}{2m.\,{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C\]
Correct Answer: C
Solution :
Let \[I=\int_{{}}^{{}}{\frac{({{x}^{5m-1}}\,+2{{x}^{4m-1}})dx}{{{x}^{6m}}{{(1+{{x}^{-m}}+{{x}^{-2m}})}^{3}}}}\] \[=\int_{{}}^{{}}{\frac{{{x}^{-(m+1)\,}}+2.x{{\,}^{-(2m+1)}}}{{{(1+{{x}^{-m}}+{{x}^{-2m}})}^{3}}}dx}\] Put \[(1+{{x}^{-m}}+{{x}^{-2m}})=t\] \[I=\frac{-1}{m}\int_{{}}^{{}}{\frac{dt}{{{t}^{3}}}\,=\frac{1}{2m{{t}^{2}}}+C=\,\frac{{{x}^{4m}}}{2m{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C}\]You need to login to perform this action.
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