Question

$ \int{\frac{1}{{{x}^{6}}+{{x}^{4}}}dx} $ is equal to

• 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

Answer 1

Answer: (C)

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 $