If ∫01 e−xdx1+ex=loge⁡(1+e)+K , then find the value of k is

I=∫01 e−xdx1+ex=∫01 dxex1+ex Put ex=z or exdx=dz or dx=dzex=dzz∴ I=∫1e dzz2(1+z)=∫1e 11+z−z−1z2dz=log(1+z)−log⁡z−1z1e=log(1+e)−log⁡e−1e−(log2−log1−1)=log(1+e)−1e−log2∴ K=−1e+log2