Question Number 68675 by Rio Michael last updated on 14/Sep/19
![Express in partial fraction f(x) ≡ ((2x^3 + x + 2)/((x^2 +1)(x+1)(x−2))) x ≠ −1,2 Hence or otherwise show that ∫_0 ^1 f(x) dx = −(1/(12))[ 13ln 2 + π]](https://www.tinkutara.com/question/Q68675.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:{Express}\:{in}\:{partial}\:{fraction}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:{f}\left({x}\right)\:\equiv\:\frac{\mathrm{2}{x}^{\mathrm{3}} \:+\:{x}\:+\:\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)}\:{x}\:\neq\:−\mathrm{1},\mathrm{2} \\ $$$${Hence}\:{or}\:{otherwise}\:\:{show}\:{that}\:\: \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right)\:{dx}\:=\:−\frac{\mathrm{1}}{\mathrm{12}}\left[\:\mathrm{13}{ln}\:\mathrm{2}\:+\:\pi\right] \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 14/Sep/19
![f(x)=((2x^3 +x+2)/((x+1)(x−2)(x^2 +1))) =(a/(x+1))+(b/(x−2)) +((cx+d)/(x^2 +1)) a =lim_(x→−1) (x+1)f(x) =((−1)/((−3)(2))) =(1/6) b =lim_(x→2) (x−2)f(x) =((16+4)/(3.5)) =((20)/(15)) =(4/3) lim_(x→+∞) xf(x)=2 =a+b +c ⇒c=2−a−b =2−(1/6)−(4/3) =((12−1−8)/6) =(1/2) ⇒f(x)=(1/(6(x+1))) +(4/(3(x−2))) +(((1/2)x+d)/(x^2 +1)) f(0)=−1 =(1/6)−(2/3) +(1/2)+d =((1−4+3)/6) +d =d⇒d=−1 ⇒ f(x) =(1/(6(x+1))) +(4/(3(x−2))) +(1/2)((x−2)/(x^(2 ) +1)) ∫_0 ^1 f(x)dx =(1/6)[ln∣x+1∣]_0 ^1 +(4/3)[ln∣x−2∣]_0 ^1 +(1/4)[ln(x^2 +1)]_0 ^1 −[arctanx]_0 ^1 =(1/6)ln(2)−(4/3)ln(2) +(1/4)ln(2)−(π/4) =((1/6)−(4/3)+(1/4))ln(2)−(π/4) =((2−16+3)/(12))ln(2)−(π/4) ⇒ ∫_0 ^1 f(x)dx =−((11)/(12))ln(2)−(π/4)](https://www.tinkutara.com/question/Q68684.png)
$${f}\left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{3}} \:+{x}+\mathrm{2}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:=\frac{{a}}{{x}+\mathrm{1}}+\frac{{b}}{{x}−\mathrm{2}}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){f}\left({x}\right)\:=\frac{−\mathrm{1}}{\left(−\mathrm{3}\right)\left(\mathrm{2}\right)}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${b}\:={lim}_{{x}\rightarrow\mathrm{2}} \left({x}−\mathrm{2}\right){f}\left({x}\right)\:=\frac{\mathrm{16}+\mathrm{4}}{\mathrm{3}.\mathrm{5}}\:=\frac{\mathrm{20}}{\mathrm{15}}\:=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xf}\left({x}\right)=\mathrm{2}\:={a}+{b}\:+{c}\:\Rightarrow{c}=\mathrm{2}−{a}−{b}\:=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{12}−\mathrm{1}−\mathrm{8}}{\mathrm{6}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{6}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{4}}{\mathrm{3}\left({x}−\mathrm{2}\right)}\:+\frac{\frac{\mathrm{1}}{\mathrm{2}}{x}+{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${f}\left(\mathrm{0}\right)=−\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{2}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{2}}+{d}\:=\frac{\mathrm{1}−\mathrm{4}+\mathrm{3}}{\mathrm{6}}\:+{d}\:={d}\Rightarrow{d}=−\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{6}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{4}}{\mathrm{3}\left({x}−\mathrm{2}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}}\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}\:} +\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{f}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{6}}\left[{ln}\mid{x}+\mathrm{1}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:+\frac{\mathrm{4}}{\mathrm{3}}\left[{ln}\mid{x}−\mathrm{2}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\left[{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{6}}{ln}\left(\mathrm{2}\right)−\frac{\mathrm{4}}{\mathrm{3}}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right){ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}}\:=\frac{\mathrm{2}−\mathrm{16}+\mathrm{3}}{\mathrm{12}}{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}}\:\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:=−\frac{\mathrm{11}}{\mathrm{12}}{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}} \\ $$
Commented by mathmax by abdo last updated on 14/Sep/19
$${there}\:{is}\:{a}\:{error}\:{in}\:{the}\:{question}! \\ $$
Commented by Rio Michael last updated on 15/Sep/19
$${anyway}\:{thanks}\:{alot}\:{guys}\:{i}\:{found}\:{my}\:{error} \\ $$