A) \[\frac{{{x}^{5m}}}{2m{{\left( {{x}^{2m}}+{{x}^{m}}+1 \right)}^{2}}}\]
B) \[\frac{{{x}^{4m}}}{2m{{\left( {{x}^{2m}}+{{x}^{m}}+1 \right)}^{2}}}\]
C) \[\frac{2m\left( {{x}^{5m}}+{{x}^{4m}} \right)}{{{\left( {{x}^{2m}}+{{x}^{m}}+1 \right)}^{2}}}\]
D) \[\frac{\left( {{x}^{5m}}-{{x}^{4m}} \right)}{2m{{\left( {{x}^{2m}}+{{x}^{m}}+1 \right)}^{2}}}\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{{{x}^{5m-1}}+2{{x}^{4m-1}}}{{{({{x}^{2m}}+{{x}^{m}}+1)}^{3}}}}dx\] \[=\int_{{}}^{{}}{\frac{{{x}^{5m-1}}+2{{x}^{4m-1}}}{{{x}^{6m}}{{(1+{{x}^{-m}}+{{x}^{-2m}})}^{3}}}}dx\] \[=\int_{{}}^{{}}{\frac{{{x}^{-m-1}}+2{{x}^{-2m-1}}}{{{(1+{{x}^{-m}}+{{x}^{-2m}})}^{3}}}}dx\] Put\[t=1+{{x}^{-m}}+{{x}^{-2m}}+{{x}^{-2m}}\] \[\therefore \]\[\frac{dt}{dx}=-m{{x}^{-m-1}}-2m{{x}^{-2m-1}}\] \[\Rightarrow \]\[\frac{dt}{-m}=({{x}^{-m-1}}+2{{x}^{-2m-1}})dx\] \[\therefore \]\[\int_{{}}^{{}}{\frac{x{{5}^{m-1}}+2{{x}^{4m-1}}}{{{({{x}^{2m}}+{{x}^{m}}+1)}^{3}}}dx\frac{1}{-m}\int_{{}}^{{}}{{{t}^{-3}}}dt=\frac{1}{2m{{t}^{2}}}+C}\] \[=\frac{1}{2m(1+{{x}^{-m}}+{{x}^{-2m}})}+C\] \[=\frac{{{x}^{4m}}}{2m{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}+C\] \[\therefore \]\[f(x)=\frac{{{x}^{4m}}}{2m{{({{x}^{2m}}+{{x}^{m}}+1)}^{2}}}\]
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