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Question Number 49941 by turbo msup by abdo last updated on 12/Dec/18

calculate ∫_0 ^1 e^(−x) ln(1+x)dx

calculate∫01e−xln(1+x)dx

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18

∫_0 ^1 e^(−x) ln(1+x)dx≈5×(0.2×0.2)

∫01e−xln(1+x)dx≈5×(0.2×0.2)

Commented by Abdo msup. last updated on 16/Dec/18

for ∣x∣<1 we hsve ln^′ (1+x)=(1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ ln(1+x)=Σ_(n=0) ^∞ (((−1)^n )/(n+1))x^(n+1) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n ⇒ ∫_0 ^1 e^(−x) ln(1+x)dx =∫_0 ^1 e^(−x) (Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n )dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) ∫_0 ^1 x^n e^(−x) dx by parts u =x^n and v^′ =e^(−x) A_n =∫_0 ^1 x^n e^(−x) dx =[−x^n e^(−x) ]_0 ^1 +∫_0 ^1 n x^(n−1) e^(−x) dx =−(1/e) +n A_(n−1) ⇒ A_n =nA_(n−1) −(1/e) ...be continued...

for∣x∣<1wehsveln′(1+x)=11+x=∑n=0∞(−1)nxn⇒ln(1+x)=∑n=0∞(−1)nn+1xn+1=∑n=1∞(−1)n−1nxn⇒∫01e−xln(1+x)dx=∫01e−x(∑n=1∞(−1)n−1nxn)dx=∑n=1∞(−1)n−1n∫01xne−xdxbypartsu=xnandv′=e−xAn=∫01xne−xdx=[−xne−x]01+∫01nxn−1e−xdx=−1e+nAn−1⇒An=nAn−1−1e...becontinued...

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