If A=∫01et1+tdt then ∫01etln(1+t)dt=

A=∫01et1+tdt now ∫01ln1+t.etdt =ln(1+t.et)01-∫0111+tetdt =eln2-0-A

If A=∫01et1+tdt then ∫01etln(1+t)dt=

A=∫01et1+tdt now ∫01ln1+t.etdt =ln(1+t.et)01-∫0111+tetdt =eln2-0-A

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If $A = \int_{0}^{1} \frac{e^{t}}{1 + t} d t t h e n \int_{0}^{1} e^{t} \ln (1 + t) d t =$

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$e \ln 2 - A$

$e \ln 2 + A$

$A \ln 2$

$A e \ln 2$

answer is A.

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$A = \int_{0}^{1} \frac{e^{t}}{1 + t} d t n o w \int_{0}^{1} \ln (1 + t) . e^{t} d t = (\ln (1 + t) . e^{t})_{0}^{1} - \int_{0}^{1} \frac{1}{1 + t} e^{t} d t = (e \ln 2 - 0) - A$

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If A=∫01et1+tdt then ∫01etln(1+t)dt=

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If $A = \int_{0}^{1} \frac{e^{t}}{1 + t} d t t h e n \int_{0}^{1} e^{t} \ln (1 + t) d t =$