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Question Number 200159 by Calculusboy last updated on 15/Nov/23

Commented by 0670322918 last updated on 15/Nov/23

∫^1 _0 ((x^3 −3x^2 +3x−1)/(x^4 +4x^3 +6x^2 +4x+1))dx= ∫_0 ^1 (((x−1)^3 )/((x+1)^4 ))dx=∫_0 ^1 (((x+1−2)^3 )/((x+1)^4 ))dx= ∫_0 ^1 (((x+1)^3 −6(x+1)^2 +12(x+1)−8)/((x+1)^4 ))dx ∫_0 ^1 ((1/(x+1))−(6/((x+1)^2 ))+((12)/((x+1)^3 ))−(8/((x+1)^4 )))dx= [ln(x+1)+(6/((x+1)))−(6/((x+1)^2 ))+(8/(3(x+1)^3 ))]_0 ^1 = ln(2)+3−(3/2)+(1/3)−6+6−(8/3)= ln(2)+(3/2)−(7/3)=((ln(64)−5)/6) ∫_0 ^1 ((x^3 −3x^2 +3x−1)/(x^4 +4x^3 +6x^2 +4x+1))dx=((ln(64)−5)/6)

∫01x3−3x2+3x−1x4+4x3+6x2+4x+1dx=∫10(x−1)3(x+1)4dx=∫01(x+1−2)3(x+1)4dx=∫10(x+1)3−6(x+1)2+12(x+1)−8(x+1)4dx∫10(1x+1−6(x+1)2+12(x+1)3−8(x+1)4)dx=Missing \left or extra \rightln(2)+3−32+13−6+6−83=ln(2)+32−73=ln(64)−56∫10x3−3x2+3x−1x4+4x3+6x2+4x+1dx=ln(64)−56

Commented by Calculusboy last updated on 15/Nov/23

thanks sir

thankssir

Answered by Frix last updated on 15/Nov/23

∫_0 ^1 (((x−1)^3 )/((x+1)^4 ))dx =^(t=x+1) ∫(((t−2)^3 )/t^4 )dt= =∫_1 ^2 (t^(−1) −6t^(−2) +12t^(−3) −8t^(−4) )dt= =[ln t +6t^(−1) −6t^(−2) +(8/3)t^(−3) ]_1 ^2 = =−(5/6)+ln 2

∫10(x−1)3(x+1)4dx=t=x+1∫(t−2)3t4dt==∫21(t−1−6t−2+12t−3−8t−4)dt==[lnt+6t−1−6t−2+83t−3]12==−56+ln2

Commented by Calculusboy last updated on 15/Nov/23

thanks sir

thankssir

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