A) \[-\frac{1}{3{{x}^{3}}}+\frac{1}{x}+\cos \text{e}{{\text{c}}^{-1}}x+C\]
B) \[-\frac{1}{3{{x}^{3}}}+\frac{1}{x}+{{\cot }^{-1}}x+C\]
C) \[-\frac{1}{3{{x}^{3}}}+\frac{1}{x}+ta{{n}^{-1}}x+C\]
D) None of the above
Correct Answer: C
Solution :
Let\[I=\int{\frac{1}{{{x}^{6}}+{{x}^{4}}}dx=\int{\frac{1}{{{x}^{4}}({{x}^{2}}+1)}dx}}\] On putting\[x=\tan \theta \]and\[dx={{\sec }^{2}}\theta \,\,d\theta \], we get \[I=\int{\frac{1}{{{\tan }^{4}}\theta (1+{{\tan }^{2}}\theta )}{{\sec }^{2}}\theta \,d\theta }\] \[=\int{{{\cot }^{4}}\theta \,\,d\theta }\] \[=\int{{{\cot }^{2}}\theta (\cos \text{e}{{\text{c}}^{2}}\theta -1)}d\theta \] \[=\int{{{\cot }^{2}}\theta }\cos \text{e}{{\text{c}}^{2}}\theta d\theta -\int{{{\cot }^{2}}\theta d\theta }\] \[=-\int{{{\cot }^{2}}\theta }d(\cot \theta )-\int{(\cos \text{e}{{\text{c}}^{2}}\theta -1)}d\theta \] \[=-\frac{{{\cot }^{3}}\theta }{3}+\cot \theta +\theta +C\] \[=-\frac{1}{3{{x}^{3}}}+\frac{1}{x}+{{\tan }^{-1}}+C\]
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