Science:Math Exam Resources/Courses/MATH101/April 2012/Question 03 (c)

MATH101 April 2012

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Question 03 (c)
Evaluate $\int {\frac {4x+8}{(x-2)(x^{2}+4)}}\,dx$

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
Use partial fractions.

Hint 2
Since the denominator can not be simplified any further you will need a polynomial of degree one in the enumerator of the ${\frac {1}{x^{2}+4}}$ term.

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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 want to write the integrand using partial fractions: ${\begin{aligned}{\frac {{\color {red}4}x+{\color {blue}8}}{(x-2)(x^{2}+4)}}&={\frac {A}{x-2}}+{\frac {Bx+C}{x^{2}+4}}\\&={\frac {A(x^{2}+4)+(Bx+C)(x-2)}{(x-2)(x^{2}+4)}}\\&={\frac {Ax^{2}{\color {blue}+4A}+Bx^{2}{\color {red}+C}x{\color {red}-2B}x{\color {blue}-2C}}{(x-2)(x^{2}+4)}}\\&={\frac {(A+B)x^{2}+{\color {red}(-2B+C)}x+{\color {blue}4A-2C}}{(x-2)(x^{2}+4)}}\end{aligned}}$ Comparing coefficients we get the equations: ${\begin{aligned}0&=A+B\\\color {red}4&\color {red}=-2B+C\\\color {blue}8&\color {blue}=4A-2C\end{aligned}}$ Now we solve: $A=-B$, which gives $4=2A+C$, adding twice this equation to the third one we get $16=8A$ so $A=2$. This means $B=-2$ and $C=0$. So we have ${\begin{aligned}\int {\frac {4x+8}{(x-2)(x^{2}+4)}}dx&=2\int {\frac {1}{x-2}}dx-\int {\frac {2x}{x^{2}+4}}dx\\&=2\ln |x-2|-\ln |x^{2}+4|+c\end{aligned}}$

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