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Question Number 118145 by bemath last updated on 15/Oct/20
∫ (dx/((x+1)^2 (x^2 +1))) ?
∫dx(x+1)2(x2+1)?
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
Decompositionfractional1(x+1)2(x2+1)=Ax+1+B(x+1)2+Cx+Dx2+1⇒1=A(x+1)+B(x2+1)+(Cx+D)(x+1)2putx=−1⇒1=2B;B=12putx=0⇒1=A+12+D;A+D=12putx=−2⇒1=−A+52+D−2C;−A+D−2C=−32putx=1⇒1=2A+1+4C+4D;A+2C+2D=0wegetA=1;C=0;D=−12.sointegralcanbewrittenasI=∫dxx+1+∫dx2(x+1)2−∫dx2(x2+1)I=ln∣x+1∣−12(x+1)−12arctan(x)+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
∫dx(x+1)2(x2+1)dx∫12x(x2+1)−12x(x+1)2dx=12∫1x−xx2+1−12∫1(x+1)(1x−1(x+1))=12log(x)−14log(x2+1)−12log(x)+12log(x+1)−12(x+1)=12(log(x+1)−1x+1−12log(x2+1))+C=12(log(x+1x2+1)−1x+1)+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
1(x+1)2(x2+1)=ax+bx2+2x+1+cx+dx2+1⇔(a+c)x3++(b+2c+d)x2+(a+c+2d)x+b+d≡1⇔{a+c=0b+2c+d=0a+c+2d=0b+d=1⇔{d=0,b=1c=−1/2,a=1/2⇒∫dx(x+1)2(x2+1)=∫(x+22(x+1)2−x2(x2+1))dx12∫(x+1(x+1)2+1(x+1)2−xx2+1)==12[∫dxx+1+∫dx(x+1)2−12∫d(x2+1)x2+1]=12ln∣x+1∣−12(x+1)−14ln∣x2+1∣+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
complexmethodI=∫dx(x+1)2(x−i)(x+i)=∫dx(x+1x−i)2(x−i)3(x+i)letx+1x−i=t⇒x+1=tx−it⇒(1−t)x=−it−1⇒x=−it−11−t=it+1t−1⇒dxdt=i(t−1)−it−1(t−1)2=−i−1(t−1)2alsox−i=it+1t−1−i=it+1−it+it−1=1+it−1andx+i=it+1t−1+i=it+1+it−it−1=2it+1−it−1⇒I=∫−i−1(t−1)2(1+it−1)3(2it+1−it−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)dtbut∫(t−1)2(2it+1−i)dt∫t2−2t+12i(t+1−i2i)dt=12i∫t2−2t+1t+αdt(α=1−i2i)=12i∫t(t+α)−αt−2t+1t+αdt=12it22−α+22i∫tt+αdt+12iln(t+α)=t24i−α+22i∫t+α−αt+αdt+12iln(t+α)=t24i−α+22it+α2+2α2iln(t+α)+12iln(t+α)+Cwitht=x+1x−iwegetI=14i(x+1x−i)2−α+22i(x+1x−i)+12i(α+1)2ln(x+1x−i+α)+C

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