Find ∫ ln [e ^ (x) + 1] / e ^ (x) DX

Find ∫ ln [e ^ (x) + 1] / e ^ (x) DX

Let e ^ x = t, then x = LNT
∫ln[e^(x)+1]/e^(x)dx =∫ln[t+1]/t d(lnt)
=∫ln[t+1]/t^2 dt
=- ∫ ln [T + 1] d (1 / T) and integral step by step
= - ln[t+1]/t + ∫1/t d(ln[t+1])
= - ln[t+1]/t + ∫(1/t)(1/(t+1)) dt
= - ln[t+1]/t + ∫1/t dt - ∫1/(t+1) dt
= - ln[t+1]/t + lnt - ln(t+1)
If t = e ^ x is carried in, the following result will be obtained:
Original formula = - ln [e ^ x + 1] / e ^ x + X - ln (e ^ x + 1)

Elden Ring Premiere

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