Remove the modulus on the actual interval · source reference

Printed question

Without the use of a calculator, show that

$$\int_{-2}^{2}\frac{|x-2|}{x^2-2x+4}dx=p\ln3+q\pi,$$

where $p$ and $q$ are exact real constants to be determined. [6]

[2023/HCI/Prelim/I/4]

Printed working

$$\begin{aligned}\int_{-2}^2\frac{|x-2|}{x^2-2x+4}dx&=\int_{-2}^2\frac{2-x}{x^2-2x+4}dx\\&=-\frac12\int_{-2}^2\frac{2x-2}{x^2-2x+4}dx+\int_{-2}^2\frac{dx}{x^2-2x+4}\\&=-\frac12[\ln(x^2-2x+4)]_{-2}^2+\int_{-2}^2\frac{dx}{(x-1)^2+(\sqrt3)^2}\\&=-\frac12[\ln4-\ln12]+\frac1{\sqrt3}\left[\tan^{-1}\frac{x-1}{\sqrt3}\right]_{-2}^2\\&=-\frac12\ln\frac13+\frac1{\sqrt3}\tan^{-1}\frac1{\sqrt3}-\frac1{\sqrt3}\tan^{-1}(-\sqrt3)\\&=\frac12\ln3+\frac1{\sqrt3}\frac\pi6-\frac1{\sqrt3}\left(-\frac\pi3\right)\\&=\frac12\ln3+\frac1{\sqrt3}\left(\frac\pi3+\frac\pi6\right)=\frac12\ln3+\frac\pi{2\sqrt3}.\end{aligned}$$

Hence $p=\frac12$, $q=\frac1{2\sqrt3}$ (shown).

Teacher notes

$x^2-2x+4=(x-1)^2+3>0$ for all $x\in\mathbb R$.

$$|x-2|=\begin{cases}x-2,&x\ge2,\\2-x,&x<2.\end{cases}$$

Therefore $|x-2|=2-x$ for $-2\le x\le2$.

Use $\int f'/f=\ln|f|+c$ with $f=x^2-2x+4>0$; MF27 $\int dx/(a^2+x^2)=a^{-1}\tan^{-1}(x/a)+c$; and the not-in-MF27 composite version with $f=x-1$, $f'=1$, $a=\sqrt3$.

Printed answer key

$p=1/2,q=1/(2\sqrt3)$.

Source diagrams in native form