Question Number 75098 by Rio Michael last updated on 07/Dec/19
$$\mathrm{the}\:\mathrm{function}\:\mathrm{f}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by}\:{f}\left({x}\right)\:=\:\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\left.{a}\right)\:\mathrm{Express}\:\mathrm{f}\:\mathrm{into}\:\mathrm{partial}\:\mathrm{fraction} \\ $$$$\mathrm{b}.\mathrm{show}\:\mathrm{that}\:\int_{\mathrm{3}} ^{\mathrm{5}} {f}\left({x}\right)\:{dx}\:=\:{ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$
Answered by peter frank last updated on 07/Dec/19
![y=(2/((x−1)(x+1))) 2=(A/(x−1))+(B/(x+1)) 2=A(x+1)+B(x−1) A=1 B=−1 y=∫((1/(x−1))−(1/(x+1)))dy y^′ =[ln(((x−1)/(x+1)))]^5 _3 y^′ =ln((2/3))−ln((1/2))=ln((4/3))](Q75103.png)
$${y}=\frac{\mathrm{2}}{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)} \\ $$$$\mathrm{2}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{B}}{{x}+\mathrm{1}} \\ $$$$\mathrm{2}={A}\left({x}+\mathrm{1}\right)+{B}\left({x}−\mathrm{1}\right) \\ $$$${A}=\mathrm{1}\:\:\:\:\:{B}=−\mathrm{1} \\ $$$${y}=\int\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dy} \\ $$$${y}^{'} =\left[{ln}\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)\underset{\mathrm{3}} {\right]}^{\mathrm{5}} \\ $$$${y}^{'} ={ln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)={ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$$$ \\ $$
Commented by Rio Michael last updated on 07/Dec/19
$${thanks}\:{sir} \\ $$