Science:Math Exam Resources/Courses/MATH101/April 2010/Question 03 (a)

MATH101 April 2010

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Question 03 (a)
Full-Solution Problem. Justify your answer and show all your work. Simplification of the answer is not required. Evaluate the following integral: $\int {\frac {x^{2}-9}{x(x^{2}+9)}}dx.$

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 fractions

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Solution 1

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 look for an expression in partial fractions of the form: ${\frac {x^{2}-9}{x(x^{2}+9)}}={\frac {A}{x}}+{\frac {Bx+C}{x^{2}+9}}$ Which translates to: $x^{2}-9=Ax^{2}+9A+Bx^{2}+Cx$ From this, we get the following system of equations: ${\begin{aligned}A+B&=1\\9A&=-9\\C&=0\end{aligned}}$ And so $A=-1$ and $B=2$. Then ${\frac {x^{2}-9}{x(x^{2}+9)}}={\frac {-1}{x}}+{\frac {2x}{x^{2}+9}}$ Now we compute the integral ${\begin{aligned}\int {\frac {x^{2}-9}{x(x^{2}+9)}}dx&=\int {\frac {-dx}{x}}+\int {\frac {2xdx}{x^{2}+9}}\\&=-\ln |x|+\ln |x^{2}+9|+C\end{aligned}}$ Where for the last equality we use the substitution $u=x^{2}+9$.

Solution 2

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 want to evaluate the integral $\int {\frac {x^{2}-9}{x(x^{2}+9)}}\,dx.$ As a first step, we split up the integral: $\int {\frac {x^{2}-9}{x(x^{2}+9)}}\,dx=\int {\frac {x^{2}}{x(x^{2}+9)}}\,dx-\int {\frac {9}{x(x^{2}+9)}}\,dx$ We can evaluate the first integral by canceling a factor of x, and then substitute u= x^2+9, du = 2dx: ${\begin{aligned}\int {\frac {x^{2}}{x(x^{2}+9)}}\,dx&=\int {\frac {x}{x^{2}+9}}\,dx=\int {\frac {1}{2u}}\,du\\&={\frac {1}{2}}\ln |u|+C={\frac {1}{2}}\ln(x^{2}+9)+C\end{aligned}}$ The second integral can be evaluated using a partial fraction decomposition ${\frac {-9}{x(x^{2}+9)}}={\frac {A}{x}}+{\frac {Bx}{x^{2}+9}}+{\frac {C}{x^{2}+9}},$ which we can solve for $-9=A(x^{2}+9)+Bx^{2}+Cx.$ When x = -1, we get -9 = 10A+B-C, when x=1, we get -9 = 10A+B+C. Subtracting the two equations gives C = 0. Finally, when x = 2, we get -9 = 13A + 4B. So 10A + B = 13 A+ 4B, which yields A = -B. Thus -9 = 10A-A = 9A, so that A = -1 and B = 1. Therefore $\int {\frac {-9}{x(x^{2}+9)}}\,dx=\int {\frac {-1}{x}}\,dx+\int {\frac {x}{x^{2}+9}}\,dx=-\ln |x|+C+\int {\frac {x}{x^{2}+9}}\,dx.$ We evaluated the last integral earlier in this problem and found $\int {\frac {x}{x^{2}+9}}\,dx={\frac {1}{2}}\ln(x^{2}+9).$ So $\int {\frac {-9}{x(x^{2}+9)}}\,dx={\frac {1}{2}}\ln(x^{2}+9)-\ln |x|+C,$ and therefore the final answer is $\int {\frac {x^{2}-9}{x(x^{2}+9)}}\,dx=\ln(x^{2}+9)-\ln |x|+C.$

Solution 3

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 want to evaluate the integral $\int {\frac {x^{2}-9}{x(x^{2}+9)}}\,dx.$ This can be rewritten as ${\begin{aligned}\int {\frac {2x^{2}-(x^{2}+9)}{x(x^{2}+9)}}\,dx&=\int {\frac {2x^{2}}{x(x^{2}+9)}}\,dx-\int {\frac {x^{2}+9}{x(x^{2}+9)}}\,dx\\&=\int {\frac {2x}{x^{2}+9}}\,dx-\int {\frac {1}{x}}\,dx.\end{aligned}}$ We evaluate the first integral using the substitution u = x^2+9, du = 2xdx, the second integral is just ln|x| + C. Hence ${\begin{aligned}\int {\frac {2x^{2}-(x^{2}+9)}{x(x^{2}+9)}}\,dx&=\int {\frac {1}{u}}\,du-\ln |x|-C\\&=\ln |u|-\ln |x|-C\\&=\ln |x^{2}+9|-\ln |x|-C\end{aligned}}$ Note: Since C is a constant, we can write -C or +D, for D=-C. Either solution is fine.

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