Find $\int \frac{1+e^x}{1-e^x}dx$ by substitution

Question

We can multiply and divide by $e^x$ to get the desired form $\int f(g(x))g'(x)dx$ and use the substitution rule in a straightforward way.

I'm reading Spivak and he says that there's an alternative and preferable way of evaluating this integral:

if we write $u=e^x$, $x=\log{u}$, $dx=\frac1udu$, then $\int \frac{1+e^x}{1-e^x}dx$ becomes $\int \frac{1+u}{1-u}\frac1udu$

I don't understand what happens between $x=\log{u}$ and $dx=\frac1udu$. It looks like he takes the the derivative of both $x$ and $log{x}$ and appends $dx$ and $du$. Why does he take the derivative? How and why does he append $dx$ and $du$?

I'm pretty sure Spivak goes over this method in detail in his chapter on integration techniques (C19 if I remember corrrectly). Also, take a look at this previous answer of mine. — peek-a-boo, Apr 16 '21 at 16:19

score 0 · Answer 1 · answered Apr 16 '21 at 16:25

Since $u=e^x$, we can take natural log on both sides $$\ln u = x$$ Now, continue by differentiating both sides w.r.t. $x$, $$\frac1u\frac{\text du}{\text dx}=1$$ which then implies $$\frac{\text du}{u}=\text dx$$

The derivative is taken because we are changing the integrand from $x$ to $u$, so we need to change the differential from $\text dx$ to $\text du$. The rest is mere substitution of $u$ and $\text du$.

Hope this helps. Ask anything if not clear :)

score 0 · Accepted Answer · edited Apr 20 '21 at 05:18

0

When $u=e^x$, taking log of both sides , we get $x=\log u$ , now differentiating both sides , we get

$dx=\frac{1}{u}du$ ,then use the value of $x$ and $dx$ in the required integral.

edited Apr 20 '21 at 05:18

Star Bright

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answered Apr 16 '21 at 16:28

My problem was that I did't think of $\frac{du}{dx}$ as a fraction – super.t Apr 18 '21 at 19:45
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@super1 That's not how it works. – J.G. Apr 20 '21 at 05:55

score 0 · Answer 3 · answered Apr 16 '21 at 16:37

The secondary path for taking that particular derivative is to begin directly with $u=e^x$. Then $du=e^xdx$ which in turn means that $du=udx\to\frac1udu=dx$. But this is just a confirmation that $x=\log u\to dx=\frac1udu$.

In general, the method of substitution for integrals always involves correlating the differential quantities. If we simply substitute $y=2x$ in $\int 4x^2dx$ to get $\int y^2dx$ our integral is still valid (if we keep the connection of $y$ with $dx$), but we have hidden the connection between $y$ and $dx$ and it is much more likely that a mistake will occur. If we make that same substitution and directly replace $dx$ by $dy$ then we have lost the connection and our integral $\int y^2dy$ does not align with our original.

Find $\int \frac{1+e^x}{1-e^x}dx$ by substitution

3 Answers3