Integral $ \int e^{x} \sqrt{e^{x} - 1} dx $

Question

I want to determinate the following integral $\int e^{x} \sqrt{e^{x} - 1} dx$

My try and steps were as follow

$$ \int e^{x} \sqrt{e^{x} - 1} dx $$

let $ u = \sqrt{e^{x} - 1} $ and $ v' = e^{x} $

$$ \to e^{x} \sqrt{e^{x} - 1} - \int \frac{e^{x}}{2\sqrt{e^{x}-1}} e^{x} dx$$

let $ u = e^{x} $ and $ v' = \frac{e^{x}}{2\sqrt{e^{x}-1}} $

$$ \to e^{x} \sqrt{e^{x} - 1} - (e^{x} \sqrt{e^{x} - 1} - \int e^{x}\sqrt{e^{x}-1} dx ) \\ = e^{x} \sqrt{e^{x} - 1} - e^{x} \sqrt{e^{x} - 1} + \int e^{x}\sqrt{e^{x}-1} dx \\ = 0 + \int e^{x}\sqrt{e^{x}-1} dx $$

What am I missing or doing wrong? Is it not allowed to change the $u$ & $v'$ when doing the 2nd time partition?

Answer 1

Hint You don't need integration by parts for this problem. Let $u=e^x-1$.

Answer 2

You are overcomplicating it. Look at this quickly:

$$ \int e^{x} \sqrt{e^{x} - 1} dx $$

Note that:

$$\frac{d}{dx}e^x=e^x$$

Well, jolly gosh! We've basically got $e^x$ in that square root! Let's use u-substitution. The inside of the root is $u$:

$$u=e^{x} - 1$$ $$du=e^{x} dx$$

Rewrite the equation as follows:

$$ \int u^{\frac{1}{2}} du $$

Hence, the integral of that would be:

$$\frac{2}{3}u^{\frac{3}{2}}$$

Substituting $u$ back:

$$\frac{2}{3}(e^{x} - 1)^{\frac{3}{2}}$$

For future, try using WolframAlpha for basic integrals. Cheers! -Shahar

Answer 3

$$ \int e^{x} \sqrt{e^{x} - 1} dx $$

Make these substitution: $$u=e^{x} - 1$$ $$du=e^{x} dx$$

Rewrite it like this:

$$ \int u^{\frac{1}{2}} du $$

Take the integral:

$$\frac{2}{3}u^{\frac{3}{2}}$$

Finally you get:

$$\frac{2}{3}(e^{x} - 1)^{\frac{3}{2}}$$