Question Number 29162 by abdo imad last updated on 04/Feb/18
$${find}\:\:{find}\:{I}=\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\:\:\frac{\mid{x}−\mathrm{2}\mid}{\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} }{dx}\:. \\ $$
Commented by abdo imad last updated on 06/Feb/18
![the chasles relation give I=∫_1 ^2 ((−x+2)/((x^2 −4x)^2 ))dx +∫_2 ^3 ((x−2)/((x^2 −4x)^2 ))dx ∫_1 ^2 ((−x +2)/((x^2 −4x)^2 ))dx=−(1/2)∫_1 ^2 ((2x−4)/((x^2 −4x)^2 ))dx =(1/2)[ (1/(x^2 −4x))]_1 ^2 =(1/2)(−(1/4) +(1/3))=(1/(24)) and ∫_2 ^3 ((x−2)/((x^2 −4x)^2 ))dx=(1/2) ∫_1 ^3 ((2x−4)/((x^2 −4x)^2 ))dx =−(1/2)[ (1/(x^2 −4x))]_1 ^3 =−(1/2)( −(1/3) +(1/3))=0 ⇒ I=(1/(24)) .](https://www.tinkutara.com/question/Q29304.png)
$${the}\:{chasles}\:{relation}\:{give}\:{I}=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{−{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} }{dx}\:+\int_{\mathrm{2}} ^{\mathrm{3}} \:\frac{{x}−\mathrm{2}}{\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} }{dx} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{−{x}\:+\mathrm{2}}{\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} }{dx}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{2}{x}−\mathrm{4}}{\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}\right]_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{24}}\:\:{and} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{3}} \:\:\frac{{x}−\mathrm{2}}{\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\frac{\mathrm{2}{x}−\mathrm{4}}{\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}\right]_{\mathrm{1}} ^{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\:−\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0}\:\:\Rightarrow\:{I}=\frac{\mathrm{1}}{\mathrm{24}}\:. \\ $$