Science:Math Exam Resources/Courses/MATH101/April 2015/Question 04 (b)

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MATH101 April 2015

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Question 04 (b)
Calculate $\int {\frac {1}{x^{4}+x^{2}}}dx.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use partial fraction decomposition.

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Solution
Using partial fraction decomposition, we can write the integrand as ${\frac {1}{x^{4}+x^{2}}}={\frac {1}{x^{2}(x^{2}+1)}}={\frac {A}{x}}+{\frac {B}{x^{2}}}+{\frac {Cx+D}{x^{2}+1}}$ for some constants $A,B,C,D$ . We find that $A=0,B=1,C=0,D=-1$ , which implies that $\int {\frac {1}{x^{4}+x^{2}}}\,dx=\int {\frac {1}{x^{2}}}\,dx-\int {\frac {1}{1+x^{2}}}\,dx.$ The first integral on the right-hand side of the equation can be easily evaluated: $\int {\frac {1}{x^{2}}}\,dx=-{\frac {1}{x}}+C.$ On the other hand, to compute the second integral, we can use the substitution $x=\tan u$ . Then $1+x^{2}=1+\tan ^{2}u=\sec ^{2}u$ and $dx=\sec ^{2}u\,du$ , which gives $\int {\frac {1}{1+x^{2}}}\,dx=\int {\frac {1}{\sec ^{2}u}}\sec ^{2}u\,du=\int 1\,du=u+C=\arctan x+C.$ Combining these, we obtain $\int {\frac {1}{x^{4}+x^{2}}}dx=\color {blue}-{\frac {1}{x}}-\arctan x+C.$