If `k=int_(0)^(1) (e^(t))/(1+t)dt`, then `int_(0)^(1) e^(t)log_(e )(1+t)dt `is equal to

Options

A. k
B. 2k
C. e In 2-k
D. none of these

Answer

C

Explanation

`k=underset(0)overset(1)int (e^(t))/(1+t)dt=underset(0)overset(1)int e^(t)(1)/(underset(I" "II" ")(1+t))dt`
`rArr k=[e^(t)log(1+t)]_(0)^(1)-underset(0)overset(1)int underset(II)e^(t) log(underset(I" ")(1+1))dt`
`rArr k=e log_(e )2-underset(0)overset(1)int e^(t) log(1+t)dt`
`rArr underset(0)overset(1)int e^(t) log(1+t)dt=e" In "2-k`