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Question Number 119335 by physicstutes last updated on 23/Oct/20

 Express f(x) = (1/((x−1)^2 (x^2 +1)))  into partial fractions.  hence evaluate I = ∫_0 ^4  f(x) dx

Expressf(x)=1(x1)2(x2+1)intopartialfractions.henceevaluateI=40f(x)dx

Answered by floor(10²Eta[1]) last updated on 23/Oct/20

(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   =(A+C)x^3 −(A−B+2C+D)x^2 +(A+C−2D)x−A+B+D  ⇒A+C=0  A−B+2C+D=0  A+C−2D=0  −A+B+D=1  ⇒D=0⇒C=1/2⇒A=−1/2⇒B=1/2  (1/((x−1)^2 (x^2 +1)))=((−1)/(2(x−1)))+(1/(2(x−1)^2 ))+(x/(2(x^2 +1)))

1(x1)2(x2+1)=Ax1+B(x1)2+Cx+Dx2+11=A(x1)(x2+1)+B(x2+1)+(Cx+D)(x1)2=(A+C)x3(AB+2C+D)x2+(A+C2D)xA+B+DA+C=0AB+2C+D=0A+C2D=0A+B+D=1D=0C=1/2A=1/2B=1/21(x1)2(x2+1)=12(x1)+12(x1)2+x2(x2+1)

Answered by mathmax by abdo last updated on 24/Oct/20

f(x)=(1/((x−1)^2 (x^2 +1))) ⇒f(x)=(a/(x−1))+(b/((x−1)^2 )) +((cx+d)/(x^2 +1))  b=(1/2) ,lim_(x→+∞) xf(x)=0 =a+c ⇒c=−a ⇒  f(x)=(a/(x−1)) +(1/(2(x−1)^2 )) +((−ax+d)/(x^2 +1))  f(0)=1 =−a+(1/2) +d ⇒−a+d=(1/2) ⇒a−d=−(1/2)  f(2)=(1/5)=a +(1/2)+((−2a+d)/5) ⇒1=5a+(5/2)−2a+d  ⇒  3a+d=1−(5/2)=−(3/2)  we have a=d−(1/2) ⇒3(d−(1/2))+d=−(3/2)  ⇒4d=0 ⇒d=0 ⇒a=−(1/2) ⇒  f(x)=−(1/(2(x−1))) +(1/(2(x−1)^2 ))+(x/(2(x^2 +1)))  so  ∫_0 ^4 f(x)dx =−(1/2)∫_0 ^4  (dx/(x−1)) +(1/2)∫_0 ^4  (dx/((x−1)^2 )) +(1/4)∫_0 ^4  ((2x)/(x^2 +1))dx  =−(1/2)[ln∣x−1∣]_0 ^4 −(1/2)[(1/(x−1))]_0 ^4 +(1/4)[ln(x^2 +1)]_0 ^4   =−(1/2)ln(3)−(1/2)((1/3)+1)+(1/4)ln(17)  =−((ln3)/2)−(2/3) +((ln17)/4)

f(x)=1(x1)2(x2+1)f(x)=ax1+b(x1)2+cx+dx2+1b=12,limx+xf(x)=0=a+cc=af(x)=ax1+12(x1)2+ax+dx2+1f(0)=1=a+12+da+d=12ad=12f(2)=15=a+12+2a+d51=5a+522a+d3a+d=152=32wehavea=d123(d12)+d=324d=0d=0a=12f(x)=12(x1)+12(x1)2+x2(x2+1)so04f(x)dx=1204dxx1+1204dx(x1)2+14042xx2+1dx=12[lnx1]0412[1x1]04+14[ln(x2+1)]04=12ln(3)12(13+1)+14ln(17)=ln3223+ln174

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