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Question Number 118145 by bemath last updated on 15/Oct/20

∫ (dx/((x+1)^2 (x^2 +1))) ?

$$\int\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:? \\ $$

Commented by bobhans last updated on 15/Oct/20

Decomposition fractional (1/((x+1)^2 (x^2 +1))) = (A/(x+1))+(B/((x+1)^2 )) +((Cx+D)/(x^2 +1)) ⇒ 1=A(x+1)+B(x^2 +1)+(Cx+D)(x+1)^2 put x=−1⇒1=2B; B=(1/2) put x=0⇒1=A+(1/2)+D ; A+D=(1/2) put x=−2⇒1=−A+(5/2)+D−2C; −A+D−2C=−(3/2) put x=1⇒1=2A+1+4C+4D; A+2C+2D=0 we get A=1; C=0 ; D=−(1/2). so integral can be written as I = ∫(dx/(x+1))+∫ (dx/(2(x+1)^2 ))−∫ (dx/(2(x^2 +1))) I= ln ∣x+1∣ −(1/(2(x+1)))−(1/2)arc tan (x) + c

$$\mathrm{Decomposition}\:\mathrm{fractional} \\ $$$$\:\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{\mathrm{A}}{\mathrm{x}+\mathrm{1}}+\frac{\mathrm{B}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{Cx}+\mathrm{D}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{1}=\mathrm{A}\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{B}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\left(\mathrm{Cx}+\mathrm{D}\right)\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{put}\:\mathrm{x}=−\mathrm{1}\Rightarrow\mathrm{1}=\mathrm{2B};\:\mathrm{B}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{put}\:\mathrm{x}=\mathrm{0}\Rightarrow\mathrm{1}=\mathrm{A}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{D}\:;\:\mathrm{A}+\mathrm{D}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{put}\:\mathrm{x}=−\mathrm{2}\Rightarrow\mathrm{1}=−\mathrm{A}+\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{D}−\mathrm{2C};\:−\mathrm{A}+\mathrm{D}−\mathrm{2C}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{put}\:\mathrm{x}=\mathrm{1}\Rightarrow\mathrm{1}=\mathrm{2A}+\mathrm{1}+\mathrm{4C}+\mathrm{4D};\:\mathrm{A}+\mathrm{2C}+\mathrm{2D}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{A}=\mathrm{1};\:\mathrm{C}=\mathrm{0}\:;\:\mathrm{D}=−\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$$\mathrm{so}\:\mathrm{integral}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\: \\ $$$$\mathrm{I}\:=\:\int\frac{\mathrm{dx}}{\mathrm{x}+\mathrm{1}}+\int\:\frac{\mathrm{dx}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }−\int\:\frac{\mathrm{dx}}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\mathrm{I}=\:\mathrm{ln}\:\mid\mathrm{x}+\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arc}\:\mathrm{tan}\:\left(\mathrm{x}\right)\:+\:\mathrm{c}\: \\ $$

Answered by Dwaipayan Shikari last updated on 15/Oct/20

∫(dx/((x+1)^2 (x^2 +1)))dx ∫(1/(2x(x^2 +1)))−(1/(2x(x+1)^2 ))dx =(1/2)∫(1/x)−(x/(x^2 +1))−(1/2)∫(1/((x+1)))((1/x)−(1/((x+1)))) =(1/2)log(x)−(1/4)log(x^2 +1)−(1/2)log(x)+(1/2)log(x+1)−(1/(2(x+1))) =(1/2)(log(x+1)−(1/(x+1))−(1/2)log(x^2 +1))+C =(1/2)(log(((x+1)/( (√(x^2 +1)))))−(1/(x+1)))+C

$$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{2}{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{x}}−\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{log}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{4}}{log}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}{log}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}{log}\left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({log}\left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}{log}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({log}\left(\frac{{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\right)−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)+{C} \\ $$

Answered by 1549442205PVT last updated on 15/Oct/20

(1/((x+1)^2 (x^2 +1)))=((ax+b)/(x^2 +2x+1))+((cx+d)/(x^2 +1)) ⇔(a+c)x^3 ++(b+2c+d)x^2 +(a+c+2d)x+b+d≡1 ⇔ { ((a+c=0)),((b+2c+d=0)),((a+c+2d=0)),((b+d=1)) :} ⇔ { ((d=0,b=1)),((c=−1/2,a=1/2)) :} ⇒∫ (dx/((x+1)^2 (x^2 +1))) =∫(((x+2)/(2(x+1)^2 ))−(x/(2(x^2 +1))))dx (1/2)∫(((x+1)/((x+1)^2 ))+(1/((x+1)^2 ))−(x/(x^2 +1)))= =(1/2)[∫(dx/(x+1))+∫(dx/((x+1)^2 ))−(1/2)∫((d(x^2 +1))/(x^2 +1))] =(1/2)ln∣x+1∣−(1/(2(x+1)))−(1/4)ln∣x^2 +1∣+C

$$\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}}+\frac{\mathrm{cx}+\mathrm{d}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Leftrightarrow\left(\mathrm{a}+\mathrm{c}\right)\mathrm{x}^{\mathrm{3}} ++\left(\mathrm{b}+\mathrm{2c}+\mathrm{d}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{a}+\mathrm{c}+\mathrm{2d}\right)\mathrm{x}+\mathrm{b}+\mathrm{d}\equiv\mathrm{1} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{a}+\mathrm{c}=\mathrm{0}}\\{\mathrm{b}+\mathrm{2c}+\mathrm{d}=\mathrm{0}}\\{\mathrm{a}+\mathrm{c}+\mathrm{2d}=\mathrm{0}}\\{\mathrm{b}+\mathrm{d}=\mathrm{1}}\end{cases}\:\Leftrightarrow\begin{cases}{\mathrm{d}=\mathrm{0},\mathrm{b}=\mathrm{1}}\\{\mathrm{c}=−\mathrm{1}/\mathrm{2},\mathrm{a}=\mathrm{1}/\mathrm{2}}\end{cases} \\ $$$$\Rightarrow\int\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:=\int\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{x}}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\right)\mathrm{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{x}+\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\int\frac{\mathrm{dx}}{\mathrm{x}+\mathrm{1}}+\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{x}^{\mathrm{2}} +\mathrm{1}\mid+\mathrm{C} \\ $$

Answered by Bird last updated on 16/Oct/20

complex method I =∫ (dx/((x+1)^2 (x−i)(x+i))) =∫ (dx/((((x+1)/(x−i)))^2 (x−i)^3 (x+i))) let ((x+1)/(x−i))=t ⇒x+1=tx−it ⇒ (1−t)x =−it−1 ⇒x =((−it−1)/(1−t)) =((it+1)/(t−1)) ⇒(dx/dt) =((i(t−1)−it−1)/((t−1)^2 )) =((−i−1)/((t−1)^2 )) also x−i=((it+1)/(t−1))−i =((it+1−it+i)/(t−1)) =((1+i)/(t−1)) and x+i =((it+1)/(t−1))+i =((it+1+it−i)/(t−1)) =((2it+1−i)/(t−1)) ⇒ I =∫ ((−i−1)/((t−1)^2 (((1+i)/(t−1)))^3 (((2it+1−i)/(t−1)))))dt =(−i−1)∫ (((t−1)^4 )/((t−1)^2 (1+i)^3 (2it+1−i))) =−(1/((1+i)^2 ))∫ (((t−1)^2 )/((2it+1−i)))dt but ∫ (((t−1)^2 )/((2it+1−i)))dt∫ ((t^2 −2t+1)/(2i(t+((1−i)/(2i)))))dt =(1/(2i))∫ ((t^2 −2t+1)/(t+α))dt (α=((1−i)/(2i))) =(1/(2i))∫ ((t(t+α)−αt−2t+1)/(t+α))dt =(1/(2i))(t^2 /2) −((α+2)/(2i))∫ (t/(t+α))dt +(1/(2i))ln(t+α) =(t^2 /(4i))−((α+2)/(2i))∫ ((t+α−α)/(t+α))dt+(1/(2i))ln(t+α) =(t^2 /(4i))−((α+2)/(2i))t +((α^2 +2α)/(2i))ln(t+α) +(1/(2i))ln(t+α) +C with t=((x+1)/(x−i)) we get I =(1/(4i))(((x+1)/(x−i)))^2 −((α+2)/(2i))(((x+1)/(x−i))) +(1/(2i))(α+1)^2 ln(((x+1)/(x−i)) +α) +C

$${complex}\:{method} \\ $$$${I}\:=\int\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−{i}\right)\left({x}+{i}\right)} \\ $$$$=\int\:\:\:\frac{{dx}}{\left(\frac{{x}+\mathrm{1}}{{x}−{i}}\right)^{\mathrm{2}} \left({x}−{i}\right)^{\mathrm{3}} \left({x}+{i}\right)} \\ $$$${let}\:\frac{{x}+\mathrm{1}}{{x}−{i}}={t}\:\Rightarrow{x}+\mathrm{1}={tx}−{it}\:\Rightarrow \\ $$$$\left(\mathrm{1}−{t}\right){x}\:=−{it}−\mathrm{1}\:\Rightarrow{x}\:=\frac{−{it}−\mathrm{1}}{\mathrm{1}−{t}} \\ $$$$=\frac{{it}+\mathrm{1}}{{t}−\mathrm{1}}\:\Rightarrow\frac{{dx}}{{dt}}\:=\frac{{i}\left({t}−\mathrm{1}\right)−{it}−\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{−{i}−\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:{also}\:{x}−{i}=\frac{{it}+\mathrm{1}}{{t}−\mathrm{1}}−{i} \\ $$$$=\frac{{it}+\mathrm{1}−{it}+{i}}{{t}−\mathrm{1}}\:=\frac{\mathrm{1}+{i}}{{t}−\mathrm{1}}\:{and} \\ $$$${x}+{i}\:=\frac{{it}+\mathrm{1}}{{t}−\mathrm{1}}+{i}\:=\frac{{it}+\mathrm{1}+{it}−{i}}{{t}−\mathrm{1}} \\ $$$$=\frac{\mathrm{2}{it}+\mathrm{1}−{i}}{{t}−\mathrm{1}}\:\Rightarrow \\ $$$${I}\:=\int\:\:\frac{−{i}−\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}+{i}}{{t}−\mathrm{1}}\right)^{\mathrm{3}} \left(\frac{\mathrm{2}{it}+\mathrm{1}−{i}}{{t}−\mathrm{1}}\right)}{dt} \\ $$$$=\left(−{i}−\mathrm{1}\right)\int\:\:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+{i}\right)^{\mathrm{3}} \left(\mathrm{2}{it}+\mathrm{1}−{i}\right)} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{\mathrm{2}} }\int\:\:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{2}{it}+\mathrm{1}−{i}\right)}{dt}\:{but} \\ $$$$\int\:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{2}{it}+\mathrm{1}−{i}\right)}{dt}\int\:\:\frac{{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}}{\mathrm{2}{i}\left({t}+\frac{\mathrm{1}−{i}}{\mathrm{2}{i}}\right)}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int\:\frac{{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}}{{t}+\alpha}{dt}\:\:\:\:\:\left(\alpha=\frac{\mathrm{1}−{i}}{\mathrm{2}{i}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int\:\frac{{t}\left({t}+\alpha\right)−\alpha{t}−\mathrm{2}{t}+\mathrm{1}}{{t}+\alpha}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:−\frac{\alpha+\mathrm{2}}{\mathrm{2}{i}}\int\:\:\frac{{t}}{{t}+\alpha}{dt}\:+\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left({t}+\alpha\right) \\ $$$$=\frac{{t}^{\mathrm{2}} }{\mathrm{4}{i}}−\frac{\alpha+\mathrm{2}}{\mathrm{2}{i}}\int\:\frac{{t}+\alpha−\alpha}{{t}+\alpha}{dt}+\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left({t}+\alpha\right) \\ $$$$=\frac{{t}^{\mathrm{2}} }{\mathrm{4}{i}}−\frac{\alpha+\mathrm{2}}{\mathrm{2}{i}}{t}\:+\frac{\alpha^{\mathrm{2}} +\mathrm{2}\alpha}{\mathrm{2}{i}}{ln}\left({t}+\alpha\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left({t}+\alpha\right)\:+{C} \\ $$$${with}\:{t}=\frac{{x}+\mathrm{1}}{{x}−{i}}\:{we}\:{get} \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\left(\frac{{x}+\mathrm{1}}{{x}−{i}}\right)^{\mathrm{2}} −\frac{\alpha+\mathrm{2}}{\mathrm{2}{i}}\left(\frac{{x}+\mathrm{1}}{{x}−{i}}\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\alpha+\mathrm{1}\right)^{\mathrm{2}} {ln}\left(\frac{{x}+\mathrm{1}}{{x}−{i}}\:+\alpha\right)\:+{C} \\ $$

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