Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 68675 by Rio Michael last updated on 14/Sep/19

$$$$$$Expressinpartialfraction$$$$f\left(x\right)\equiv \frac{2x^3+x+2}{(x^2+1)(x+1)(x-2)}x\ne -1,2$$$$Henceorotherwiseshowthat$$$${\int }_0^{1}f\left(x\right)dx=-\frac{1}{12}[13ln2+\pi ]$$$$$$

Commented by mathmax by abdo last updated on 14/Sep/19

$$f\left(x\right)=\frac{2x^3+x+2}{(x+1)(x-2)(x^2+1)}=\frac{a}{x+1}+\frac{b}{x-2}+\frac{cx+d}{x^2+1}$$$$a={lim}_{x\to -1}(x+1)f\left(x\right)=\frac{-1}{(-3)\left(2\right)}=\frac{1}{6}$$$$b={lim}_{x\to 2}(x-2)f\left(x\right)=\frac{16+4}{3.5}=\frac{20}{15}=\frac{4}{3}$$$${lim}_{x\to +\mathrm{\infty }}xf\left(x\right)=2=a+b+c\Rightarrow c=2-a-b=2-\frac{1}{6}-\frac{4}{3}$$$$=\frac{12-1-8}{6}=\frac{1}{2}\Rightarrow f\left(x\right)=\frac{1}{6(x+1)}+\frac{4}{3(x-2)}+\frac{\frac{1}{2}x+d}{x^2+1}$$$$f\left(0\right)=-1=\frac{1}{6}-\frac{2}{3}+\frac{1}{2}+d=\frac{1-4+3}{6}+d=d\Rightarrow d=-1\Rightarrow$$$$f\left(x\right)=\frac{1}{6(x+1)}+\frac{4}{3(x-2)}+\frac{1}{2}\frac{x-2}{x^2+1}$$$${\int }_0^{1}f\left(x\right)dx=\frac{1}{6}{[ln\mid x+1\mid ]}_0^1+\frac{4}{3}{[ln\mid x-2\mid ]}_0^1+\frac{1}{4}{\left[ln(x^2+1)\right]}_0^1$$$$-{\left[arctanx\right]}_0^1=\frac{1}{6}ln\left(2\right)-\frac{4}{3}ln\left(2\right)+\frac{1}{4}ln\left(2\right)-\frac{\pi }{4}$$$$=(\frac{1}{6}-\frac{4}{3}+\frac{1}{4})ln\left(2\right)-\frac{\pi }{4}=\frac{2-16+3}{12}ln\left(2\right)-\frac{\pi }{4}\Rightarrow$$$${\int }_0^{1}f\left(x\right)dx=-\frac{11}{12}ln\left(2\right)-\frac{\pi }{4}$$

Commented by mathmax by abdo last updated on 14/Sep/19

$$thereisaerrorinthequestion!$$

Commented by Rio Michael last updated on 15/Sep/19

$$anywaythanksalotguysifoundmyerror$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com