Integral utter confusion with substition and dx/du

Question

I need to find the indefinite integral I = $$\int e^x (1+e^x)^{\frac{1}{2}}$$

by using a proper substition method. I tried it on https://www.integral-calculator.com and it gave the following explanation

Now this is where it gets confusing, when substituting $u = (1+e^x)$ I would suspect we got $$\int e^x \sqrt{(1+e^x)} = \int e^x\sqrt{u} $$ But that's not what the screenshot says.

It says that the original problem equals $\int \sqrt{u} du$ and also that $dx = e^{-x}$. Now where does this $e^{-x}$ come from? The antiderivite of $e^x$ is still $e^x$. And the antiderivative of $e^x+1$ is $e^x+x$ so where does that negative number come from and why is it used?

So those are my questions. 1. Why is $$\int e^x \sqrt{(1+e^x)} = \int \sqrt{u} du $$ 2. Where does the $dx = e^{-x}$ come from 3. WHy is it even necessary?

Answer 1

$(1)$ Because when you make the substitution $\color{purple}{u = e^x+1}$, taking the derivative of $u$ yields $\frac{du}{dx} = e^x$. Hence, $\color{blue}{e^xdx = \frac{du}{dx}dx = du}$. The integral can therefore be rewritten:

$$\int \color{blue}{e^x}\sqrt{\color{purple}{e^x+1}} \color{blue}{dx} \implies \int\sqrt{\color{purple}{u}}\color{blue}{du}$$

$(2)$ In the previous part, note that we get $\frac{du}{dx} = e^x$. This can be rewritten as $\frac{du}{e^x} = dx$, and $e^x$ in the denominator can be written as $e^{-x}$ in the numerator, so $dx = e^{-x}du$.

$(3)$ It’s certainly not necessary. It makes the integral easier to evaluate.

Answer 2

If $e^{x}-1 =u \rightarrow e^xdx=du$ (I suggest writing it this way) so the integral after the substitution looks like this:
$\int (e^x \sqrt{e^x-1})dx=\int(\sqrt{u})du=\cfrac{2u^{3/2}}{3} + C=\cfrac{2(e^x-1)^{3/2}}{3}+C$

Answer 3

The way I learnt it more then sixty years ago.

$$I=\int e^x \sqrt{e^x+1}\,dx$$ $$e^x+1=u\implies x=\log(u-1) \implies dx=\frac{du}{u-1}$$ making $$I=\int (u-1)\, \sqrt u \,\frac{du}{u-1}=\int \sqrt u \,du=\cdots$$

Answer 4

Another approach:

\begin{equation} I = \int e^x \sqrt{e^x + 1}\:dx \end{equation}

Here let $u^2 = e^x + 1$. Employing Implicit Differentiation we see

\begin{equation} \frac{d}{dx}\left[u^2\right] = \frac{d}{dx}\left[e^x + 1 \right] \rightarrow 2u \frac{du}{dx} = e^x \rightarrow dx = \frac{2u}{e^x}\:du = \frac{2u}{u^2 - 1}\:du \end{equation}

Thus:

\begin{align} I &= \int e^x \sqrt{e^x + 1}\:dx = \int \left(u^2 - 1\right)\sqrt{u^2}\frac{2u}{u^2 - 1}\:du \\ &= \int 2u^2 \:du = \frac{u^3}{3} + C = \frac{1}{3}\left(e^x + 1 \right)^\frac{3}{2} + C \end{align}

Where $C$ is the constant of integration.