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Question Number 47248 by maxmathsup by imad last updated on 06/Nov/18
calculate ∫_(−1) ^1 ((ln(x+2))/((x+4)^2 −1))dx
calculate∫−11ln(x+2)(x+4)2−1dx
Commented by maxmathsup by imad last updated on 16/Nov/18
let A = ∫_(−1) ^1 ((ln(x+2))/((x+4)^2 −1))dx ⇒ A =_(x+2=t) ∫_1 ^3 ((ln(t))/((t+2)^2 −1))dt =∫_1 ^3 ((ln(t))/((t+1)(t+3)))dt =(1/2)∫_1 ^3 {(1/(t+1)) −(1/(t+3))}ln(t)dt =(1/2) ∫_1 ^3 ((ln(t))/(t+1))dt −(1/2) ∫_1 ^3 ((ln(t))/(t+3))dt let determine ∫ ((ln(x))/(x+1))dx the calculator give ∫ ((ln(x))/(x+1)) dx=ln(x)ln(x+1) +L_(i 2) (−x) ⇒ ∫_1 ^3 ((ln(t))/(t+1))dt =[ln(t)ln(t+1)+L_(i2) (−t)]_1 ^3 =2ln(3)ln(2) +L_(i2) (−3) −L_(i2) (−1) also ∫ ((ln(x))/(x+3))dx =ln(x)ln((x/3)+1)+L_(i2) (−(x/3)) ⇒ ∫_1 ^3 ((ln(x))/(x+3)) =[ln(x)ln((x/3)+1)+L_(i2) (−(x/3))]_1 ^3 =ln(3)ln(2)+L_(i2) (−1)−L_(i2) (−(1/3))....
letA=∫−11ln(x+2)(x+4)2−1dx⇒A=x+2=t∫13ln(t)(t+2)2−1dt=∫13ln(t)(t+1)(t+3)dt=12∫13{1t+1−1t+3}ln(t)dt=12∫13ln(t)t+1dt−12∫13ln(t)t+3dtletdetermine∫ln(x)x+1dxthecalculatorgive∫ln(x)x+1dx=ln(x)ln(x+1)+Li2(−x)⇒∫13ln(t)t+1dt=[ln(t)ln(t+1)+Li2(−t)]13=2ln(3)ln(2)+Li2(−3)−Li2(−1)also∫ln(x)x+3dx=ln(x)ln(x3+1)+Li2(−x3)⇒∫13ln(x)x+3=[ln(x)ln(x3+1)+Li2(−x3)]13=ln(3)ln(2)+Li2(−1)−Li2(−13)….

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