Damon Stokes

2022-06-30

U-substitution for integral of 1/(1+e^x)dx. What am I doing wrong?

Here is my work, with 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 $u=\frac{1}{{e}^{z}}+1$, $\frac{du}{dz}=-{e}^{-z}dz\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}-du=\frac{1}{{e}^{z}}dz$

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

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

Here is my work, with 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 $u=\frac{1}{{e}^{z}}+1$, $\frac{du}{dz}=-{e}^{-z}dz\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}-du=\frac{1}{{e}^{z}}dz$

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

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

Arcatuert3u

Beginner2022-07-01Added 30 answers

Your answer is correct.

$-\mathrm{ln}({e}^{-z}+1)+C=-\mathrm{ln}\left[{e}^{-z}(1+{e}^{z})\right]+C=\dots $

$-\mathrm{ln}({e}^{-z}+1)+C=-\mathrm{ln}\left[{e}^{-z}(1+{e}^{z})\right]+C=\dots $

sviraju6d

Beginner2022-07-02Added 6 answers

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

$z+\text{ln}(1+{e}^{-z})==\text{ln}(1+{e}^{z})\phantom{\rule{0ex}{0ex}}{e}^{z+\text{ln}(1+{e}^{-z})}=={e}^{\text{ln}(1+{e}^{z})}\phantom{\rule{0ex}{0ex}}{e}^{z}(1+{e}^{-z})==1+{e}^{z}\phantom{\rule{0ex}{0ex}}{e}^{z}+1==1+{e}^{z}$

$z+\text{ln}(1+{e}^{-z})==\text{ln}(1+{e}^{z})\phantom{\rule{0ex}{0ex}}{e}^{z+\text{ln}(1+{e}^{-z})}=={e}^{\text{ln}(1+{e}^{z})}\phantom{\rule{0ex}{0ex}}{e}^{z}(1+{e}^{-z})==1+{e}^{z}\phantom{\rule{0ex}{0ex}}{e}^{z}+1==1+{e}^{z}$

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