A) \[\frac{{{x}^{4}}}{6{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{3}}}+C\]
B) \[\frac{{{x}^{12}}}{6{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{3}}}+C\]
C) \[\frac{{{x}^{4}}}{{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{3}}}+C\]
D) \[\frac{{{x}^{12}}}{{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{3}}}+C\]
Correct Answer: B
Solution :
\[\int{\frac{3{{x}^{13}}+2{{x}^{11}}}{{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{4}}}dx}\] |
\[\int{\frac{\left( \frac{3}{{{x}^{3}}}+\frac{2}{{{x}^{5}}} \right)dx}{\left( 2+\frac{3}{{{x}^{2}}}+\frac{1}{{{x}^{4}}} \right)}}\] |
Let \[\left( 2+\frac{3}{{{x}^{2}}}+\frac{1}{{{x}^{4}}} \right)=t\] |
\[-\frac{1}{2}\int{\frac{dt}{{{t}^{4}}}=\frac{1}{6{{t}^{3}}}+c}\] \[\Rightarrow \]\[\frac{{{x}^{12}}}{6{{(2{{x}^{4}}+3{{x}^{2}}+1)}^{3}}}+C\] |
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