That’s not what the original question was asking. The question wanted the *definite* integral from 0 to 1 of 1/(e^x + 1), not the indefinite integral. I think you missed a step.
@@d-hat-vr2002 Yes, and they didn’t do that. That’s my entire point. They only found the anti derivative and never actually used it to find the area under that curve from 0 to 1.
At 3:08 a partial fractions decomposition would be much easier for most people, compared to the un-motivated manipulation of the numerator shown.
Solution:
1 1 1
∫1/(1+e^x)*dx = ∫(1+e^x)/(1+e^x)*dx-∫e^x/(1+e^x)*dx
0 0 0
1 1
= ∫dx-∫e^x/(1+e^x)*dx =
0 0
----------------------------
Substitution for the 2nd integral: u = 1+e^x du = e^x*dx
lower limit = 1+e^0 = 1+1 = 2 upper limit = 1+e^1 = 1+e
----------------------------
1 1+e 1+e
= [x]-∫1/u*du = 1-[ln|u|] = 1-[ln(1+e)-ln(2)] = 1+ln(2)-ln(1+e)
0 2 2
= 1+ln[2/(1+e)] ≈ 0,3799
Holy headache! 1/(1 + e^x) = 1 - [e^x/(1 + e^x)]
I = x - ln(1 + e^x)
Answer: 1 + ln2 - ln(1 + e)
∫dx/(1+e^x)
=∫e^-x/(e^-x+1)
=-ln(e^-x+1) + C
That’s not what the original question was asking. The question wanted the *definite* integral from 0 to 1 of 1/(e^x + 1), not the indefinite integral.
I think you missed a step.
@@d-hat-vr2002 Yes, and they didn’t do that. That’s my entire point. They only found the anti derivative and never actually used it to find the area under that curve from 0 to 1.