Science:Math Exam Resources/Courses/MATH101/April 2012/Question 03 (c)/Solution 1

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

${\begin{aligned}\int {\frac {4x+8}{(x-2)(x^{2}+4)}}dx&=2\int {\frac {1}{x-2}}dx-\int {\frac {2x}{x^{2}+4}}dx\\&=2\ln |x-2|-\ln |x^{2}+4|+c\end{aligned}}$