Science:Math Exam Resources/Courses/MATH105/April 2017/Question 02 (a)

MATH105 April 2017

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Question 02 (a)
(a) Evaluate $\int {\frac {3x^{2}+2x+8}{4x^{2}-x^{3}}}dx$.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
This integral may look intimidating, but it can be solved by considering things one piece at a time. First, notice that $\int {\frac {3x^{2}}{4x^{2}-x^{3}}}\,$ can be computed by cancelling out the $x^{2}$ terms. Once you compute this integral, try to compute the remaining piece $\int {\frac {2x+8}{4x^{2}-x^{3}}}\,dx$. If you are stuck on this integral, the next hint will provide some guidance.

Hint 2

After following the procedure in the previous hint, we arrived at the integral $\int {\frac {2x+8}{4x^{2}-x^{3}}}\,dx.$ If the numerator had instead been 8 - 2x, we could have computed the integral by cancelling out a factor of 4-x. So we rewrite this integral as $\int {\frac {8-2x}{4x^{2}-x^{3}}}\,dx+\int {\frac {4x}{4x^{2}-x^{3}}}\,dx$ we can cancel out a factor of x in the second integral and use a partial fraction decomposition.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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 want to evaluate the integral $\int {\frac {3x^{2}+2x+8}{4x^{2}-x^{3}}}\,dx.$ The first thing we notice is that the denominator of this fraction factors into (x^2)(4 - x). We first compute the integral $\int {\frac {3x^{2}}{4x^{2}-x^{3}}}\,dx=\int {\frac {3}{4-x}}\,dx=-3\ln |4-x|+C$. It remains to compute the integral $\int {\frac {2x+8}{x^{2}(4-x)}}\,dx$. The easiest way to compute this integral is to modify the numerator to cancel out the 4-x in the denominator. This would be easy to do if we had 8 - 2x instead of 8 + 2x in the numerator, so we rewrite the integral: $\int {\frac {8-2x}{x^{2}(4-x)}}\,dx+\int {\frac {4x}{x^{2}(4-x)}}\,dx$. In the first integral, we cancel out a factor of 4-x: $\int {\frac {2}{x^{2}}}\,dx=-2x^{-1}+C$. The second integral, on the other hand, is $\int {\frac {4}{x(4-x)}}\,dx.$ we need to use a partial fraction decomposition in order to evaluate this integral. We rewrite ${\frac {4}{x(4-x)}}={\frac {A}{x}}+{\frac {B}{4-x}}.$ If we multiply both sides by x(4-x), we get $4=A(4-x)+Bx$. We can solve this (e.g. by plugging in x=0 and x = 4) to get the solution A = 1, B = 1. Thus we have $\int {\frac {4}{x(4-x)}}\,dx=\int {\frac {1}{x}}\,dx+\int {\frac {1}{4-x}}\,dx=\ln |x|-\ln |4-x|+C$. We have now fully computed the integral, and all that remains is to add: $\int {\frac {3x^{2}+2x+8}{4x^{2}-x^{3}}}=-3\ln |4-x|-{\frac {2}{x}}+\ln |x|-\ln |4-x|+C=-4\ln |4-x|+\ln |x|-{\frac {2}{x}}+C.$ Answer: $\color {blue}-4\ln |4-x|+\ln |x|-{\frac {2}{x}}+C$