Science:Math Exam Resources/Courses/MATH105/April 2016/Question 02 (c)

MATH105 April 2016

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Question 02 (c)
Evaluate $\int {\frac {81}{x^{3}-9x^{2}}}\,dx$.

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
Factorize the denominator and use the partial factions method.

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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. Since the degree of the numerator is smaller than the denominator which is 3 here, we use partial fractions to evaluate the integral. The denominator is $x^{3}-9x^{2}=x^{2}(x-9)$, so we split the denominator as the following, ${\frac {81}{x^{3}-9x^{2}}}={\frac {A}{x^{2}}}+{\frac {B}{x}}+{\frac {D}{x-9}}.$ We find the numerator by computing the common denominator ${\frac {81}{x^{3}-9x^{2}}}={\frac {A}{x^{2}}}+{\frac {B}{x}}+{\frac {D}{x-9}}={\frac {A(x-9)+Bx(x-9)+Dx^{2}}{x^{2}(x-9)}}.$ We simplify the numerator and make it equal to $81$ to get $(B+D)x^{2}+(A-9B)x-9A=81.$ Matching the coefficient, we find the following equations for $A,B,C$: $B+D=0,A-9B=0,-9A=81$ which gives: $A=-9,B=-1,D=1$, hence $\int {\frac {81}{x^{3}-9x^{2}}}dx=-\int {\frac {9}{x^{2}}}dx-\int {\frac {1}{x}}dx+\int {\frac {1}{x-9}}dx={\frac {9}{x}}-\ln |x|+\ln |x-9|+C.$