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Question Number 43322 by Raj Singh last updated on 09/Sep/18

Commented by maxmathsup by imad last updated on 11/Sep/18

let A = ∫ (dx/((x+1)^2 (x^2 +1))) let decompose F(x) = (1/((x+1)^2 (x^2 +1))) F(x) = (a/(x+1)) +(b/((x+1)^2 )) +((cx +d)/(x^2 +1)) b =lim_(x→−1) (x+1)^2 F(x) =(1/2) lim_(x→+∞) xF(x) =0 =a+c ⇒c =−(1/2) ⇒F(x) =(a/(x+1)) +(1/(2(x+1)^2 )) +((−(1/2)x +d)/(x^2 +1)) F(o) =1 = a +(1/2) +d ⇒ a+d =(1/2) F(2) =(1/(45)) = (a/3) +(1/(18)) +((d−1)/5) ⇒ 1 =15a +((45)/(18)) +9d−9 ⇒ 1 =15a +(5/2) +9d−9 ⇒15a +9d =10−(5/2) =((15)/2) ⇒ 15a +9((1/2)−a)=((15)/2) ⇒6a =((15)/2) −(9/2) =3 ⇒a=(1/2) ⇒d=0 ⇒ F(x)= (1/(2(x+1))) +(1/(2(x+1)^2 )) −(x/(2(1+x^2 ))) ⇒ A = ∫ F(x)dx =(1/2)ln∣x+1∣ −(1/(2(x+1))) −(1/4)ln(1+x^2 ) +c .

$${let}\:{A}\:=\:\int\:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:{let}\:{decompose}\:{F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)\:=\:\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{b}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{cx}\:+{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${b}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${lim}_{{x}\rightarrow+\infty} \:{xF}\left({x}\right)\:=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({x}\right)\:=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{−\frac{\mathrm{1}}{\mathrm{2}}{x}\:+{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left({o}\right)\:=\mathrm{1}\:=\:{a}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+{d}\:\Rightarrow\:{a}+{d}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${F}\left(\mathrm{2}\right)\:=\frac{\mathrm{1}}{\mathrm{45}}\:=\:\frac{{a}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{18}}\:+\frac{{d}−\mathrm{1}}{\mathrm{5}}\:\Rightarrow\:\mathrm{1}\:=\mathrm{15}{a}\:+\frac{\mathrm{45}}{\mathrm{18}}\:\:+\mathrm{9}{d}−\mathrm{9}\:\Rightarrow \\ $$$$\mathrm{1}\:=\mathrm{15}{a}\:+\frac{\mathrm{5}}{\mathrm{2}}\:+\mathrm{9}{d}−\mathrm{9}\:\Rightarrow\mathrm{15}{a}\:+\mathrm{9}{d}\:=\mathrm{10}−\frac{\mathrm{5}}{\mathrm{2}}\:=\frac{\mathrm{15}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{15}{a}\:+\mathrm{9}\left(\frac{\mathrm{1}}{\mathrm{2}}−{a}\right)=\frac{\mathrm{15}}{\mathrm{2}}\:\Rightarrow\mathrm{6}{a}\:=\frac{\mathrm{15}}{\mathrm{2}}\:−\frac{\mathrm{9}}{\mathrm{2}}\:=\mathrm{3}\:\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{d}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{{x}}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${A}\:=\:\int\:{F}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}+\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+{c}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18

2)∫(dx/((x+1)^2 (x^2 +1))) ∫(a/((x+1)))dx+∫(b/((x+1)^2 ))dx+∫((cx+d)/(x^2 +1))dx now calculating a b c and d (1/((x+1)^2 (x^2 +1)))=(a/(x+1))+(b/((x+1)^2 ))+((cx+d)/(x^2 +1)) 1=a(x+1)(x^2 +1)+b(x^2 +1)+(cx+d)(x+1)^2 1=a(x^3 +x^2 +x+1)+b(x^2 +1)+(cx+d)(x^2 +2x+1) 1=a(x^3 +x^2 +x+1)+b(x^2 +1)+(cx^3 +2cx^2 +cx+dx^2 +2dx+d) 1=x^3 (a+c)+x^2 (a+b+2c+d)+x(a+c+2d)+(a+b+d) a+c==0 a+b+2c+d=0 a+c+2d=0 a+b+d=1 a+c+d=0 0+d=0 d=0 a+b=1 a+b+2c+d=0 1+2c+0=0 c=((−1)/2) a+c=0 a=−c a=(1/2) a=(1/2) a+b=1 b=(1/2) c=((−1)/2) d=0 ∫(a/(x+1))dx+∫(b/((x+1)^2 ))+∫((cx)/(x^2 +1))dx (1/2)∫(dx/(x+1))+(1/2)∫(dx/((x+1)^2 )) + ((−1)/2)∫(dx/(x^2 +1)) (1/2)ln(x+1)+(1/2)×(((x+1)^(−1) )/(−1))−(1/2)tan^(−1) (x)+c (1/2)ln(x+1)−(1/(2(x+1)))−(1/2)tan^(−1) (x)+c

$$\left.\mathrm{2}\right)\int\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\int\frac{{a}}{\left({x}+\mathrm{1}\right)}{dx}+\int\frac{{b}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}+\int\frac{{cx}+{d}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${now}\:{calculating}\:{a}\:{b}\:{c}\:{and}\:{d} \\ $$$$\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{a}}{{x}+\mathrm{1}}+\frac{{b}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{cx}+{d}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{1}={a}\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{b}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\left({cx}+{d}\right)\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{1}={a}\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+{b}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\left({cx}+{d}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\mathrm{1}={a}\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+{b}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\left(\mathrm{c}{x}^{\mathrm{3}} +\mathrm{2}{cx}^{\mathrm{2}} +{cx}+{dx}^{\mathrm{2}} +\mathrm{2}{dx}+{d}\right) \\ $$$$\mathrm{1}={x}^{\mathrm{3}} \left({a}+{c}\right)+{x}^{\mathrm{2}} \left({a}+{b}+\mathrm{2}{c}+{d}\right)+{x}\left({a}+{c}+\mathrm{2}{d}\right)+\left({a}+{b}+{d}\right) \\ $$$${a}+{c}==\mathrm{0} \\ $$$${a}+{b}+\mathrm{2}{c}+{d}=\mathrm{0} \\ $$$$ \\ $$$${a}+{c}+\mathrm{2}{d}=\mathrm{0} \\ $$$${a}+{b}+{d}=\mathrm{1} \\ $$$${a}+{c}+{d}=\mathrm{0} \\ $$$$\mathrm{0}+{d}=\mathrm{0}\:\:{d}=\mathrm{0} \\ $$$${a}+{b}=\mathrm{1} \\ $$$${a}+{b}+\mathrm{2}{c}+{d}=\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{2}{c}+\mathrm{0}=\mathrm{0} \\ $$$${c}=\frac{−\mathrm{1}}{\mathrm{2}}\:\:\:\:\:{a}+{c}=\mathrm{0}\:\:\:{a}=−{c} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:{a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:{a}+{b}=\mathrm{1}\:\:\:{b}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:{c}=\frac{−\mathrm{1}}{\mathrm{2}}\:\:\:{d}=\mathrm{0} \\ $$$$\int\frac{{a}}{{x}+\mathrm{1}}{dx}+\int\frac{{b}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+\int\frac{{cx}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\:+\:\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({x}+\mathrm{1}\right)^{−\mathrm{1}} }{−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left({x}\right)+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left({x}\right)+{c} \\ $$$$ \\ $$

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