∫ [ex/ (e2x + 2ex + 1)] dx
u = ex
du = ex dx
dx = du/ ex
ln u = ln ex
ln u = x
∫[u/ (u2 + 2u + 1)](du/ u) = ∫ du/ (u2 + 2u + 1) = ∫ du/ (u + 1)2 = [- 1/ (u + 1)] + c
∫ [ex/ (e2x + 2ex + 1)] dx = [- 1/ (ex + 1)] + c
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