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Question Number 36180 by prof Abdo imad last updated on 30/May/18

calculate ∫_0 ^1 ((ln(t))/((1+t)^2 ))dt

calculate∫01ln(t)(1+t)2dt

Commented by prof Abdo imad last updated on 30/May/18

let put I = ∫_0 ^1 ((ln(t))/((1+t)^2 ))dt let integrate by parts I =[(1−(1/(1+t)))ln(t)]_0 ^1 −∫_0 ^1 (1−(1/(1+t)))(dt/t) =0 − ∫_0 ^1 (dt/(1+t)) =−[ln∣1+t∣]_0 ^1 =−ln(2) let prove that lim_(t→0) (1−(1/(1+t)))ln(t)=0 lim_(t→0) (1−(1/(1+t)))ln(t) =lim_(t→0) ((tln(t))/(1+t)) =0 because lim_(t→0) tln(t) =0 so ★ ∫_0 ^1 ((ln(t))/((1+t)^2 ))dt =−ln(2)★

letputI=∫01ln(t)(1+t)2dtletintegratebypartsI=[(1−11+t)ln(t)]01−∫01(1−11+t)dtt=0−∫01dt1+t=−[ln∣1+t∣]01=−ln(2)letprovethatlimt→0(1−11+t)ln(t)=0limt→0(1−11+t)ln(t)=limt→0tln(t)1+t=0becauselimt→0tln(t)=0so★∫01ln(t)(1+t)2dt=−ln(2)★

Answered by sma3l2996 last updated on 30/May/18

I=∫_0 ^1 ((lnt)/((1+t)^2 ))dt by parts u=lnt⇒u′=(1/t) v′=(1/((1+t)^2 ))⇒v=−(1/(1+t)) I=[−((lnt)/(1+t))]_0 ^1 +∫_0 ^1 (dt/(t(1+t))) =lim_(t→0^+ ) ((lnt)/(1+t))+∫_0 ^1 ((1/t)−(1/(1+t)))dt =lim_(t→0^+ ) ((lnt)/(1+t))+[ln((t/(1+t)))]_0 ^1 =lim_(t→0^+ ) (((lnt)/(1+t))−ln(t))−ln2 =−lim_(x→0^+ ) ((tln(t))/(1+t))−ln2=−ln2 (because lim_(x→0^+ ) xlnx=0)

I=∫01lnt(1+t)2dtbypartsu=lnt⇒u′=1tv′=1(1+t)2⇒v=−11+tI=[−lnt1+t]01+∫01dtt(1+t)=limt→0+lnt1+t+∫01(1t−11+t)dt=limt→0+lnt1+t+[ln(t1+t)]01=limt→0+(lnt1+t−ln(t))−ln2=−limx→0+tln(t)1+t−ln2=−ln2(becauselimxlnxx→0+=0)

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