Science:Math Exam Resources/Courses/MATH101/April 2013/Question 06 (b)

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MATH101 April 2013

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Question 06 (b)
Full-Solution Problem. Justify your answers and show all your work. Simplification of numerical answers is required. Evaluate $\displaystyle \int {\frac {2x-1}{x^{2}-2x+5}}\ dx$

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
Try completing the square in the denominator, then use a substitution.

Hint 2
(alternate) Is there a simple substitution? If not, can you modify the integrand so that there is?

Hint 3
What is the derivative of the denominator? How does it compare to the numerator?

Hint 4
Don't forget the constant!

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 1

Let's complete the square in the denominator. $\displaystyle {\begin{aligned}x^{2}-2x+5=(x-1)^{2}+4\end{aligned}}$ This suggests using the substitution $\displaystyle u=x-1$, so that $\displaystyle du=dx$ and $\displaystyle 2x-1=2(x-1)+1=2u+1$. Then $\displaystyle {\begin{aligned}\int {\frac {2x-1}{x^{2}-2x+5}}\ dx&=\int {\frac {2u+1}{u^{2}+4}}\,du\\&=\int {\frac {2u}{u^{2}+4}}\ du+\int {\frac {1}{u^{2}+4}}\,du\end{aligned}}$ For the first integral, we use the substitution $\displaystyle v=u^{2}+4$, so that $\displaystyle dv=2u\ du$. $\displaystyle {\begin{aligned}\int {\frac {2u}{u^{2}+4}}\ du&=\int {\frac {1}{v}}\,dv\\&=\ln |v|+C_{1}\\&=\ln |u^{2}+4|+C_{1}\\&=\ln |(x-1)^{2}+4|+C_{1}\\&=\ln |x^{2}-2x+5|+C_{1}\end{aligned}}$ For the second integal, this looks like an $\displaystyle \arctan$ integral, but we need to scale. We use the substitution $\displaystyle 2w=u$, so that $\displaystyle 2\ dw=du$ and $\displaystyle u^{2}+4=4[(u/2)^{2}+1]=4(w^{2}+1)$. Then $\displaystyle {\begin{aligned}\int {\frac {1}{u^{2}+4}}\ du&=\int {\frac {1}{4(w^{2}+1)}}2\,dw\\&={\frac {1}{2}}\int {\frac {1}{w^{2}+1}}\,dw\\&={\frac {1}{2}}\arctan w+C_{2},\\&={\frac {1}{2}}\arctan \left({\frac {u}{2}}\right)+C_{2}\\&={\frac {1}{2}}\arctan \left({\frac {x-1}{2}}\right)+C_{2}\end{aligned}}$ Altogether, $\displaystyle {\begin{aligned}\int {\frac {2x-1}{x^{2}-2x+5}}\ dx&=\ln |x^{2}-2x+5|+{\frac {1}{2}}\arctan \left({\frac {x-1}{2}}\right)+C_{3}\end{aligned}}$ with the arbitrary constant of integration C3.

Solution 2

Note that this is already in reduced form using partial fraction decomposition. We see that the numerator is almost the derivative of the denominator. We can modify it to be exactly the derivative by subtracting 1, then adding 1 to make up for it. Then using a simple substitution in the first integral and completing the square in the second, $\displaystyle {\begin{aligned}\int {\frac {2x-1}{x^{2}-2x+5}}\ dx&=\int {\frac {2x-2}{x^{2}-2x+5}}\ dx+\int {\frac {1}{x^{2}-2x+5}}\ dx\\&=\ln |x^{2}-2x+5|+\int {\frac {1}{(x-1)^{2}+2^{2}}}\ dx\\&=\ln |x^{2}-2x+5|+{\frac {1}{2}}\arctan \left({\frac {x-1}{2}}\right)+C\\\end{aligned}}$

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