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Given-that-f-x-2-x-2-1-a-Express-f-x-in-partial-fraction-b-Evaluate-3-5-f-x-dx-

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Question Number 65166 by Rio Michael last updated on 25/Jul/19
Given that f(x) = (2/(x^2 −1)) a) Express f(x) in partial fraction. b.Evaluate ∫_3 ^5 f (x) dx
Giventhatf(x)=2x2−1a)Expressf(x)inpartialfraction.b.Evaluate∫35f(x)dx
Commented by mathmax by abdo last updated on 26/Jul/19
a) we have f(x) =(2/((x−1)(x+1))) =(1/(x−1))−(1/(x+1)) b) ∫_3 ^5 f(x)dx =∫_3 ^5 ((1/(x−1))−(1/(x+1)))dx =[ln∣x−1∣−ln∣x+1∣]_3 ^5 =[ln∣((x−1)/(x+1))∣]_3 ^5 =ln((4/6))−ln((2/4)) =ln((2/3))−ln((1/2)) =ln(2)−ln(3)+ln2 =2ln2 −ln(3) .
a)wehavef(x)=2(x−1)(x+1)=1x−1−1x+1b)∫35f(x)dx=∫35(1x−1−1x+1)dx=[ln∣x−1∣−ln∣x+1∣]35=[ln∣x−1x+1∣]35=ln(46)−ln(24)=ln(23)−ln(12)=ln(2)−ln(3)+ln2=2ln2−ln(3).
Answered by mr W last updated on 25/Jul/19
f(x)=(2/(x^2 −1))=(1/(x−1))−(1/(x+1)) ∫_3 ^5 f(x)dx =∫_3 ^5 ((1/(x−1))−(1/(x+1)))dx =[ln ((x−1)/(x+1))]_3 ^5 =ln ((5−1)/(5+1))−ln ((3−1)/(3+1)) =ln (2/3)−ln (1/2) =ln (4/3)
f(x)=2x2−1=1x−1−1x+1∫35f(x)dx=∫35(1x−1−1x+1)dx=[lnx−1x+1]35=ln5−15+1−ln3−13+1=ln23−ln12=ln43
Commented by Rio Michael last updated on 25/Jul/19
thanks
thanks
Commented by Rio Michael last updated on 25/Jul/19
any explanation on how ln((x−1)/(x+1)) s obtained
anyexplanationonhowlnx−1x+1sobtained
Commented by MJS last updated on 25/Jul/19
∫(x−1)dx=ln (x−1) ∫(x+1)dx=ln (x+1) ln a −ln b =ln (a/b) e^(ln a −ln b) =e^(ln a) ×e^(−ln a) =(e^(ln a) /e^(ln b) )=(a/b) ⇒ ⇒ ln e^(ln a −ln b) =ln a −ln b =ln (a/b)
∫(x−1)dx=ln(x−1)∫(x+1)dx=ln(x+1)lna−lnb=lnabelna−lnb=elna×e−lna=elnaelnb=ab⇒⇒lnelna−lnb=lna−lnb=lnab
Commented by mr W last updated on 25/Jul/19
∫((1/(x−1))−(1/(x+1)))dx =∫(1/(x−1))dx−∫(1/(x+1))dx =∫(1/(x−1))d(x−1)−∫(1/(x+1))d(x+1) =ln (x−1)−ln (x+1) =ln ((x−1)/(x+1))
∫(1x−1−1x+1)dx=∫1x−1dx−∫1x+1dx=∫1x−1d(x−1)−∫1x+1d(x+1)=ln(x−1)−ln(x+1)=lnx−1x+1
Commented by Rio Michael last updated on 25/Jul/19
thanks you guys so much
thanksyouguyssomuch
Answered by meme last updated on 25/Jul/19
a) f(x)=(2/((x−1)(x+1)))=(1/((x−1)))−(1/((x+1))) b)∫_3 ^5 f(x) dx=∫_3 ^5 (1/((x−1))) dx −∫_3 ^5 (1/((x+1))) dx =[ln(x−1)]_3 ^5 −[ln(x+1)]_3 ^5 =ln4−ln2−ln6+ln4 =4ln2−ln2−ln6 =3ln2−ln6 =ln(4/3)
a)f(x)=2(x−1)(x+1)=1(x−1)−1(x+1)b)∫35f(x)dx=∫351(x−1)dx−∫351(x+1)dx=[ln(x−1)]35−[ln(x+1)]35=ln4−ln2−ln6+ln4=4ln2−ln2−ln6=3ln2−ln6=ln43

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