Calculate the integral: integral from 0 to (ln 2)

Calculate the integral: $\int_0^{\ln 2} \frac{e^x-1}{\sqrt{e^x-1}} \, dx$.

We will simplify the integrand and then use substitution.

Step 1: Simplify the integrand

\[\frac{e^x-1}{\sqrt{e^x-1}} = \frac{e^x-1}{(e^x-1)^{1/2}} = (e^x-1)^{1/2} = \sqrt{e^x-1}\]

Therefore:

\[\int_0^{\ln 2} \frac{e^x-1}{\sqrt{e^x-1}} \, dx = \int_0^{\ln 2} \sqrt{e^x-1} \, dx\]

Step 2: Use substitution

Let $u = e^x - 1$. Then $du = e^x \, dx$, which means $dx = \frac{du}{u+1}$ (since $e^x = u+1$).

Step 3: Change the limits of integration

When $x=0$: $u=e^0-1=0$. When $x=\ln 2$: $u=e^{\ln 2}-1=2-1=1$.

Step 4: Rewrite the integral

\[\int_0^{\ln 2} \sqrt{e^x-1} \, dx = \int_0^1 \sqrt{u} \cdot \frac{du}{u+1} = \int_0^1 \frac{u^{1/2}}{u+1} \, du\]

Step 5: Use another substitution

Let $v=\sqrt{u}$, so $u=v^2$ and $du=2v\,dv$. When $u=0$: $v=0$; when $u=1$: $v=1$.

\[\int_0^1 \frac{u^{1/2}}{u+1} \, du = \int_0^1 \frac{v}{v^2+1} \cdot 2v \, dv = \int_0^1 \frac{2v^2}{v^2+1} \, dv\]

Step 6: Simplify the integrand

We can rewrite the integrand by polynomial division:

\[\frac{2v^2}{v^2+1} = \frac{2(v^2+1) - 2}{v^2+1} = 2 - \frac{2}{v^2+1}\]

Therefore:

\[\int_0^1 \frac{2v^2}{v^2+1} \, dv = \int_0^1 2 \, dv - \int_0^1 \frac{2}{v^2+1} \, dv\]

Step 7: Integrate

We know that $\int \frac{1}{v^2+1} \, dv = \arctan(v) + C$, so:

\[= 2[v]_0^1 - 2[\arctan(v)]_0^1\]

Step 8: Evaluate at the limits

\[= 2(1-0) - 2\left(\arctan(1)-\arctan(0)\right) = 2 - 2\left(\frac{\pi}{4}-0\right) = 2 - \frac{\pi}{2}\]

Numerical value:

Using $\frac{\pi}{2}\approx 1.5708$: $2-\frac{\pi}{2}\approx 0.4292$.

Verification of the simplification:

\[\frac{e^x-1}{\sqrt{e^x-1}} = (e^x-1)^{1-1/2} = (e^x-1)^{1/2} = \sqrt{e^x-1}\]

Geometric interpretation:

The integral $\int_0^{\ln 2} \sqrt{e^x-1} \, dx$ represents the area under the curve $y=\sqrt{e^x-1}$ from $x=0$ to $x=\ln 2$. At $x=0$: $y=\sqrt{e^0-1}=0$ (the curve starts at the origin); at $x=\ln 2$: $y=\sqrt{2-1}=1$. The function is continuous and increasing on $[0,\ln 2]$, and the result $2-\frac{\pi}{2}$ is positive, as expected for the area under a positive function.

Final answer:

\[\boxed{\int_0^{\ln 2} \frac{e^x-1}{\sqrt{e^x-1}} \, dx = 2 - \frac{\pi}{2} \approx 0.4292}\]