Question

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

Answer 1

`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`