A. ln(1 + e^2u) + C
B. 1/2 ln |1+e^u| + C
C. 1/2 tan^-1 e^u + C
D. tan^-1 e^u + C
E. 1/2tan e^2u + C
Let's use a U sub! (but I'm gonna use x instead because the problem is given in terms of u)
x=1+e^u
dx = e^u du
Integral( dx / x ) = ln(x) + c
= ln(1+e^u) + c
Confirmed with Wolfram:
http://www.wolframalpha.com/input/?i=Integral+of+%...
Did you mistype something? The answer does not appear to be present.
Let v = e^u
dv = e^u du so dv = v du and du = dv/v
Substitute to get
integral ( e^u / 1+e^u du) =
integral ( v / 1+v dv/v)
= integral (1/(1+v) dv)
= ln|1 + v| + C
= ln|1 + e^u| + C
None of the listed answers are correct.
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Verified answer
Let's use a U sub! (but I'm gonna use x instead because the problem is given in terms of u)
x=1+e^u
dx = e^u du
Integral( dx / x ) = ln(x) + c
= ln(1+e^u) + c
Confirmed with Wolfram:
http://www.wolframalpha.com/input/?i=Integral+of+%...
Did you mistype something? The answer does not appear to be present.
Let v = e^u
dv = e^u du so dv = v du and du = dv/v
Substitute to get
integral ( e^u / 1+e^u du) =
integral ( v / 1+v dv/v)
= integral (1/(1+v) dv)
= ln|1 + v| + C
= ln|1 + e^u| + C
None of the listed answers are correct.