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If lambda=int0^1(e^t)/(1+t), then int0^1...

If `lambda=int_0^1(e^t)/(1+t)`, then `int_0^1e^tlog_e(1+t)dt` is equal to

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`int_(0)^(1)e^(t)log_(e)(1+t)dt`
`=[e^(t)log_(e)(1+t)]_(0)^(1)-int_(0)^(1)1/(1+t) . e^(t)dt`
`=e log_(e)2-lamda`
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CENGAGE-DEFINITE INTEGRATION -JEE Advanced Previous Year
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