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Question Number 64762 by Lontum Hans-Sandys last updated on 21/Jul/19
Given that g(x)=(2/((1+x)(1+3x^2 )) a) express g(x) in partial fractions. b) evaluate ∫_0 ^1 g((x) dx.
$$\mathrm{Given}\:\mathrm{that}\:\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{2}}{\left(\mathrm{1}+\mathrm{x}\right)\left(\mathrm{1}+\mathrm{3x}^{\mathrm{2}} \right.} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{express}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{in}\:\mathrm{partial}\:\mathrm{fractions}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{evaluate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{g}\left(\left(\mathrm{x}\right)\:\mathrm{dx}.\right. \\ $$
Commented by mathmax by abdo last updated on 21/Jul/19
1)g(x) =(a/(x+1)) +((bx+c)/(3x^2 +1)) a =lim_(x→−1) (x+1)g(x) =(1/2) lim_(x→+∞) xg(x) =0 =a+(b/3) ⇒3a+b =0 ⇒b=−(3/2) ⇒ g(x) =(1/(2(x+1))) −(1/2) ((3x−2c)/(3x^2 +1)) g(0) =2 =(1/2) +c ⇒c =2−(1/2) =(3/2) ⇒g(x) =(1/(2(x+1)))−(1/2)((3x−3)/(3x^2 +1)) b) ∫_0 ^1 g(x)dx =(1/2) ∫_0 ^1 (dx/(x+1)) −(1/2) ∫_0 ^1 ((x−1)/((x^2 +(1/3))))dx ∫_0 ^1 (dx/(x+1)) =[ln∣x+1∣]_0 ^1 =ln(2) ∫_0 ^1 ((x−1)/(x^2 +(1/3)))dx =(1/2) ∫_0 ^1 ((2x)/(x^2 +(1/3)))dx −∫_0 ^1 (dx/(x^2 +(1/3))) =[ln(x^2 +(1/3))]_0 ^1 −∫_0 ^1 (dx/(x^2 +(1/3))) =ln((4/3))−ln((1/3))−∫_0 ^1 (dx/(x^2 +(1/3))) =2ln(2)−∫_0 ^1 (dx/(x^2 +(1/3))) but we have ∫_0 ^1 (dx/(x^2 +(1/3))) =_(x =(t/( (√3)))) =∫_0 ^(√3) (dt/(((√3)/3)(1+t^2 ))) =(√3)[ arctan(t)]_0 ^(√3) =(√3) arctan((√3))=((π(√3))/3) ⇒ ∫_0 ^1 g(x)dx =((ln(2))/2) −(1/2){2ln(2)−((π(√3))/3)} =−((ln(2))/2) +((π(√3))/6)
$$\left.\mathrm{1}\right){g}\left({x}\right)\:=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{bx}+{c}}{\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){g}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xg}\left({x}\right)\:=\mathrm{0}\:={a}+\frac{{b}}{\mathrm{3}}\:\Rightarrow\mathrm{3}{a}+{b}\:=\mathrm{0}\:\Rightarrow{b}=−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$${g}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{3}{x}−\mathrm{2}{c}}{\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${g}\left(\mathrm{0}\right)\:=\mathrm{2}\:=\frac{\mathrm{1}}{\mathrm{2}}\:+{c}\:\Rightarrow{c}\:=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow{g}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{3}{x}−\mathrm{3}}{\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\left.{b}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} {g}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}−\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}\right)}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}+\mathrm{1}}\:=\left[{ln}\mid{x}+\mathrm{1}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:={ln}\left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}}{dx}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\left[{ln}\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}}\:={ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)−{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\mathrm{2}{ln}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}}\:\:{but}\:{we}\:{have}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}}\:=_{{x}\:=\frac{{t}}{\:\sqrt{\mathrm{3}}}} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\:\frac{{dt}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\sqrt{\mathrm{3}}\left[\:{arctan}\left({t}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:=\sqrt{\mathrm{3}}\:{arctan}\left(\sqrt{\mathrm{3}}\right)=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{3}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {g}\left({x}\right){dx}\:=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}{ln}\left(\mathrm{2}\right)−\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{3}}\right\}\:=−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$ \\ $$
Answered by Tanmay chaudhury last updated on 21/Jul/19
(2/((1+x)(1+3x^2 )))=(a/(1+x))+((bx+c)/(1+3x^2 )) 2=a(1+3x^2 )+(1+x)(bx+c) 2=a+3ax^2 +bx+c+bx^2 +cx 2=x^2 (3a+b)+x(b+c)+(a+c) 3a+b=0 b+c=0 so 3a−c=0 c=3a a+c=2 3a+a=2→a=(2/4)=(1/2) and c=(3/2) b=−c b=((−3)/2) hence (2/((1+x)((1+3x^2 )))=((1/2)/(x+1))+((((−3x)/2)+(3/2))/(1+3x^2 )) ∫(2/((1+x)(1+3x^2 )))=∫((1/2)/(x+1))dx+(((−3)/2))∫((xdx)/(1+3x^2 ))+(3/2)∫(dx/(1+3x^2 )) =(1/2)∫(dx/(x+1))+((−3)/(2×6))∫((6x)/(1+3x^2 ))dx+(1/2)∫(dx/(x^2 +(1/3))) =(1/2)∫(dx/(x+1))−(1/4)∫((d(1+3x^2 ))/(1+3x^2 ))+(1/2)∫(dx/(x^2 +((√(1/3)) )^2 )) =(1/2)ln(x+1)−(1/4)ln(1+3x^2 )+(1/2)×(1/(1/( (√3))))tan^(−1) ((x/(1/( (√3))))) =(1/2)ln(x+1)−(1/4)ln(1+3x^2 )+((√3)/2)tan^(−1) (x(√3) ) so ∫_0 ^1 g(x)dx =∣(1/2)ln(x+1)−(1/4)ln(1+3x^2 )+((√3)/2)tan^(−1) (x(√3) )∣_(0=) ^1 =(1/2)ln2−(1/4)ln(4)+((√3)/2)×(π/3) =(π/(2(√3)))
$$\frac{\mathrm{2}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)}=\frac{{a}}{\mathrm{1}+{x}}+\frac{{bx}+{c}}{\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} } \\ $$$$\mathrm{2}={a}\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)+\left(\mathrm{1}+{x}\right)\left({bx}+{c}\right) \\ $$$$\mathrm{2}={a}+\mathrm{3}{ax}^{\mathrm{2}} +{bx}+{c}+{bx}^{\mathrm{2}} +{cx} \\ $$$$\mathrm{2}={x}^{\mathrm{2}} \left(\mathrm{3}{a}+{b}\right)+{x}\left({b}+{c}\right)+\left({a}+{c}\right) \\ $$$$\mathrm{3}{a}+{b}=\mathrm{0} \\ $$$${b}+{c}=\mathrm{0} \\ $$$${so}\:\mathrm{3}{a}−{c}=\mathrm{0} \\ $$$${c}=\mathrm{3}{a} \\ $$$${a}+{c}=\mathrm{2} \\ $$$$\mathrm{3}{a}+{a}=\mathrm{2}\rightarrow{a}=\frac{\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:\:{c}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${b}=−{c} \\ $$$${b}=\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$${hence}\:\frac{\mathrm{2}}{\left(\mathrm{1}+{x}\right)\left(\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)\right.}=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{x}+\mathrm{1}}+\frac{\frac{−\mathrm{3}{x}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}}{\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} } \\ $$$$\int\frac{\mathrm{2}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)}=\int\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{x}+\mathrm{1}}{dx}+\left(\frac{−\mathrm{3}}{\mathrm{2}}\right)\int\frac{{xdx}}{\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}+\mathrm{1}}+\frac{−\mathrm{3}}{\mathrm{2}×\mathrm{6}}\int\frac{\mathrm{6}{x}}{\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} }{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{d}\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{tan}^{−\mathrm{1}} \left(\frac{{x}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{3}}\:\right) \\ $$$${so}\:\int_{\mathrm{0}} ^{\mathrm{1}} {g}\left({x}\right){dx} \\ $$$$=\mid\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{3}}\:\right)\mid_{\mathrm{0}=} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{4}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\pi}{\mathrm{3}} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

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