A) \[\frac{{{x}^{4}}}{6{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{3}}}+C\]
B) \[\frac{{{x}^{4}}}{{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{3}}}+C\]
C) \[\frac{{{x}^{12}}}{6{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{3}}}+C\]
D) \[\frac{{{x}^{12}}}{{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{3}}}+C\]
Correct Answer: C
Solution :
Let\[I=\int_{{}}^{{}}{\frac{3{{x}^{13}}+2{{x}^{11}}}{{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{4}}}}dx\] \[=\int_{{}}^{{}}{\frac{\frac{3}{{{x}^{3}}}+\frac{2}{{{x}^{5}}}}{{{\left( 2+\frac{3}{{{x}^{2}}}+\frac{1}{{{x}^{4}}} \right)}^{4}}}}dx\] Now, put\[2+\frac{3}{{{x}^{2}}}+\frac{1}{{{x}^{4}}}=z\] \[\therefore \]\[\left( \frac{-6}{{{x}^{3}}}-\frac{4}{{{x}^{5}}} \right)dx=dz\] \[\Rightarrow \]\[\left( \frac{3}{{{x}^{3}}}+\frac{2}{{{x}^{5}}} \right)dx=\frac{-dz}{2}\] \[\therefore \]\[I=\frac{-1}{2}\int_{{}}^{{}}{\frac{dz}{{{z}^{4}}}=\frac{-1}{2}}\left( \frac{{{z}^{-3}}}{-3} \right)+C\] \[=\frac{1}{6{{z}^{3}}}+C=\frac{1}{6{{\left( 2+\frac{3}{{{x}^{2}}}+\frac{1}{{{x}^{4}}} \right)}^{3}}}+C\] \[=\frac{{{x}^{12}}}{6{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{3}}}+C\]
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