Science:Math Exam Resources/Courses/MATH103/April 2010/Question 04 (c)

MATH103 April 2010

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Question 04 (c)
Compute the following integrals. (Leave your answers in terms of $e$ or $\ln()$.) $I=\int _{0}^{1}{\frac {x}{(2-x)(1+2x)}}\,dx$

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
The factored bottom should make partial fractions a natural choice for a technique to attempt this problem.

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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 A and B be constants such that ${\frac {x}{(2-x)(1+2x)}}={\frac {A}{2-x}}+{\frac {B}{1+2x}}={\frac {A(1+2x)+B(2-x)}{(2-x)(1+2x)}}$ Since the denominators are the same, the numerators must be equal and hence $\displaystyle x=A(1+2x)+B(2-x)$ Now, since this equality holds for all values of x, we can plug in specific values for x to get the coefficients. At $x=2$, we have $\displaystyle 2=A(1+2(2))+B(2-(2))=5A$ and so $A={\tfrac {2}{5}}$. At $x=-{\tfrac {1}{2}}$, we have $-{\tfrac {1}{2}}=A(1+2(-{\tfrac {1}{2}}))+B(2-(-{\tfrac {1}{2}}))={\tfrac {5}{2}}B$ and so $B={\tfrac {-1}{5}}$. Hence, we have ${\begin{aligned}&\int _{0}^{1}{\frac {x}{(2-x)(1+2x)}}\,dx\\&=\int _{0}^{1}{\frac {\tfrac {2}{5}}{2-x}}\,dx+\int _{0}^{1}{\frac {\tfrac {-1}{5}}{1+2x}}\,dx\\&=\left.\left({\frac {2}{5}}\ln |2-x|(-1)-{\frac {1}{5}}\ln |1+2x|{\frac {1}{2}}\right)\right|_{0}^{1}\\&=\left.\left({\frac {-2}{5}}\ln |2-x|-{\frac {1}{10}}\ln |1+2x|\right)\right|_{0}^{1}\\&={\frac {-2}{5}}\ln |2-1|-{\frac {1}{10}}\ln |1+2(1)|-{\frac {-2}{5}}\ln |2-0|-{\frac {1}{10}}\ln |1+2(0)|\\&=-{\frac {1}{10}}\ln |3|-{\frac {-2}{5}}\ln |2|\\&=-{\frac {\ln(3)}{10}}+{\frac {2\ln(2)}{5}}\end{aligned}}.$

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