Science:Math Exam Resources/Courses/MATH101/April 2006/Question 03 (b)

MATH101 April 2006

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Question 03 (b)
Full-Solution Problems. Justify your answers and show all your work. Simplification of answers is not required. Evaluate the following integral. $\int {\frac {x^{2}+2}{x^{3}+x}}\,dx$

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
Try integrating by partial fractions.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We proceed by partial fractions. Let ${\frac {x^{2}+2}{x^{3}+x}}={\frac {x^{2}+2}{x(x^{2}+1)}}={\frac {A}{x}}+{\frac {Bx+C}{x^{2}+1}}={\frac {A(x^{2}+1)+(Bx+C)x}{x^{2}+1}}$ where above we used the fact that $x^{2}+1$ is irreducible (its discriminant is $-4<0$). Thus, we know that $x^{2}+2=A(x^{2}+1)+(Bx+C)x$ Plugging in $x=0$ gives $2=0^{2}+2=A(0^{2}+1)+(B(0)+C)(0)=A$ and so $A=2$. Plugging in $x=1$ gives $3=1^{2}+2=A(1^{2}+1)+(B(1)+C)(1)=2A+B+C=4+B+C$ and so $B+C=-1$. Plugging in $x=-1$ gives $3=(-1)^{2}+2=A((-1)^{2}+1)+(B(-1)+C)(-1)=2A+B-C=4+B-C$ and so $B-C=-1$. Adding the last two equations and dividing by 2 gives $B=-1$ and so $C=0$ Hence, we have ${\begin{aligned}\int {\frac {x^{2}+2}{x^{3}+x}}\,dx&=\int {\frac {2dx}{x}}-\int {\frac {x}{x^{2}+1}}\,dx\\&=2\ln |x|-\int {\frac {x}{x^{2}+1}}\,dx\end{aligned}}$ To solve the last integral, use substitution. Let $u=x^{2}+1$ so that $du=2xdx$ and so ${\begin{aligned}-\int {\frac {x}{x^{2}+1}}\,dx=-{\frac {1}{2}}\int {\frac {du}{u}}=-{\frac {1}{2}}\ln |u|+C=-{\frac {\ln |x^{2}+1|}{2}}+C\\\end{aligned}}$ and so ${\begin{aligned}\int {\frac {x^{2}+2}{x^{3}+x}}\,dx&=2\ln |x|-\int {\frac {x}{x^{2}+1}}\,dx\\&=2\ln |x|-{\frac {\ln |x^{2}+1|}{2}}+C\end{aligned}}$

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