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Question Number 33590 by abdo imad last updated on 19/Apr/18
let α >1 calculate f(α) = ∫_α ^(+∞) ((x^2 −x+1)/((x−1)^2 (x+1)^2 )) dx .
letα>1calculatef(α)=∫α+∞x2−x+1(x−1)2(x+1)2dx.
Commented by abdo imad last updated on 22/Apr/18
we have f(α) = ∫_α ^(+∞) ((x^2 −2x+1 +x)/((x−1)^2 (x+1)^2 ))dx =∫_α ^(+∞) (((x−1)^2 +x)/((x−1)^2 (x+1)^2 ))dx = ∫_α ^(+∞) (dx/((x+1)^2 )) +∫_α ^(+∞) (x/((x−1)^2 (x+1)^2 ))dx =[((−1)/(x+1))]_α ^(+∞) +∫_α ^(+∞) ((xdx)/((x−1)^2 (x+1)^2 )) =(1/(α+1)) + I let findI F(x) = (x/((x−1)^2 (x+1)^2 ))=(a/(x−1)) +(b/((x−1)^2 )) +(c/(x+1)) +(d/((x+1)^2 )) b= lim_(x→1) (x−1)^2 F(x) =(1/4) d =lim_(x→−1) (x+1)^2 F(x)=((−1)/4) ⇒ F(x)=(a/(x−1)) +(1/(4(x−1)^2 )) +(c/(x+1)) +((−1)/(4(x+1)^2 )) F(0) =0 =−a +(1/4) +c −(1/4) ⇒c=a ⇒ F(x) =(a/(x−1)) +(1/(4(x−1)^2 )) +(a/(x+1)) −(1/(4(x+1)^2 )) F(2)= (2/9) = a +(1/4) +(a/3) −(1/(36)) ⇒2=9a +(9/4) +3 −(1/4) ⇒ 2 =9a +5 ⇒9a =−3 ⇒ a=−(1/3) ⇒ F(x)=−(1/(3(x−1))) +(1/(4(x−1)^2 )) −(1/(3(x+1))) −(1/(4(x+1)^2 )) I =∫_α ^(+∞) ( (1/(3(1−x))) −(1/(3(1+x))))dx +(1/4) ∫_α ^(+∞) (dx/((x−1)^2 )) −(1/4) ∫_α ^(+∞) (dx/((x+1)^2 )) =[(1/3)ln∣((1−x)/(1+x))∣]_α ^(+∞) −(1/4)[ (1/(x−1))]_α ^(+∞) +(1/4)[ (1/(x+1))]_α ^(+∞) =(1/3)ln∣((1−α)/(1+α))∣ +(1/(4(α−1))) −(1/(4(α+1))) ⇒ f(α) = (3/(4(α+1))) +(1/3)ln∣((1−α)/(1+α))∣ +(1/(4(α−1))) .
wehavef(α)=∫α+∞x2−2x+1+x(x−1)2(x+1)2dx=∫α+∞(x−1)2+x(x−1)2(x+1)2dx=∫α+∞dx(x+1)2+∫α+∞x(x−1)2(x+1)2dx=[−1x+1]α+∞+∫α+∞xdx(x−1)2(x+1)2=1α+1+IletfindIF(x)=x(x−1)2(x+1)2=ax−1+b(x−1)2+cx+1+d(x+1)2b=limx→1(x−1)2F(x)=14d=limx→−1(x+1)2F(x)=−14⇒F(x)=ax−1+14(x−1)2+cx+1+−14(x+1)2F(0)=0=−a+14+c−14⇒c=a⇒F(x)=ax−1+14(x−1)2+ax+1−14(x+1)2F(2)=29=a+14+a3−136⇒2=9a+94+3−14⇒2=9a+5⇒9a=−3⇒a=−13⇒F(x)=−13(x−1)+14(x−1)2−13(x+1)−14(x+1)2I=∫α+∞(13(1−x)−13(1+x))dx+14∫α+∞dx(x−1)2−14∫α+∞dx(x+1)2=[13ln∣1−x1+x∣]α+∞−14[1x−1]α+∞+14[1x+1]α+∞=13ln∣1−α1+α∣+14(α−1)−14(α+1)⇒f(α)=34(α+1)+13ln∣1−α1+α∣+14(α−1).

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