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Here is my work, witth the right answer. I feel like every step is right, but somehow I am getting the wrong answer. How?

$$ \int \frac{1}{1+e^z}dz = \int\frac{1}{e^z(\frac{1}{e^z} + 1)}dz =\int\frac{1}{\frac{1}{e^z} + 1}\frac{1}{e^z}dz $$ subbing $u = \frac{1}{e^z} + 1$, $\frac{du}{dz} = -e^{-z}dz \implies -du = \frac{1}{e^z}dz$ so

$$\int \frac{1}{1+e^z}dz = -\int\frac{1}{u}du = -\ln(u) + C = -\ln(e^{-z}+1) + C$$

But the right answer is $z - \ln(1+e^z)+C$

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2 Answers 2

Your answer is correct. $$ -\ln \left( e^{-z}+1\right) + C = -\ln \left[ e^{-z} \left( 1 + e^z\right)\right] + C = \dots $$

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The "right answer" is merely the simplification of the answer you found.

$$ z+\text{ln}\left(1+e^{-z}\right)==\text{ln}\left(1+e^z\right)\\ e^{z+\text{ln}\left(1+e^{-z}\right)}==e^{\text{ln}\left(1+e^z\right)}\\ e^z\left(1+e^{-z}\right)==1+e^z\\ e^z+1==1+e^z $$

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