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Question Number 41516 by maxmathsup by imad last updated on 08/Aug/18

calculate ∫_0 ^1 ((ln(1+x))/((1+x)^4 )) dx

calculate∫01ln(1+x)(1+x)4dx

Commented by turbo msup by abdo last updated on 12/Aug/18

let A =∫_0 ^1 ((ln(1+x))/((1+x)^4 ))dx changement 1+x =t give A = ∫_1 ^2 ((ln(t))/t^4 )dt and by parts u^′ =t^(−4) and v=ln(t) ⇒ A =[−(1/3)t^(−3) ln(t)]_1 ^2 +∫_1 ^2 (1/3) t^(−3) (dt/t) =−(1/3)((ln(2))/8) +(1/3) ∫_1 ^2 t^(−4) dt =−((ln(2))/(24)) +(1/3)[−(1/3)t^(−3) ]_1 ^2 =−((ln(2))/(24)) −(1/9)( (1/8) −1) =−((ln(2))/(24)) −(1/9)(((−7)/8)) =(7/(72)) −((ln(2))/(24))

letA=∫01ln(1+x)(1+x)4dxchangement1+x=tgiveA=∫12ln(t)t4dtandbypartsu′=t−4andv=ln(t)⇒A=[−13t−3ln(t)]12+∫1213t−3dtt=−13ln(2)8+13∫12t−4dt=−ln(2)24+13[−13t−3]12=−ln(2)24−19(18−1)=−ln(2)24−19(−78)=772−ln(2)24

Answered by alex041103 last updated on 09/Aug/18

∫_0 ^1 ((ln(1+x))/((1+x)^4 )) dx=∫_0 ^1 ((ln(1+x))/((1+x)^4 )) d(x+1)= =∫_1 ^2 ((ln(x))/x^4 ) dx=I let x=e^u ⇒dx=e^u du ⇒I=∫_0 ^( ln(2)) (u/e^(4u) )e^u du=∫_0 ^(ln(2)) ue^(−3u) du= =−(1/3)∫_0 ^(ln(2)) ud(e^(−3u) ) We integrate by parts: −3I=ue^(−3u) ]_0 ^(ln(2)) −∫_0 ^(ln(2)) e^(−3u) du= =((ln(2))/8)−(7/(24)) ⇒I=(7/(72))−((ln(2))/(24))≈0.0683411

∫01ln(1+x)(1+x)4dx=∫01ln(1+x)(1+x)4d(x+1)==∫12ln(x)x4dx=Iletx=eu⇒dx=eudu⇒I=∫0ln(2)ue4ueudu=∫0ln(2)ue−3udu==−13∫0ln(2)ud(e−3u)Weintegratebyparts:−3I=ue−3u]0ln(2)−∫0ln(2)e−3udu==ln(2)8−724⇒I=772−ln(2)24≈0.0683411

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