Calculate the integral: integral from 0 to (ln 5)

Calculate the integral: $\int_0^{\ln 5} \frac{\sqrt{e^x-1}}{1+3e^{-x}} \, dx$.

We will use substitution to simplify this integral.

Step 1: Simplify the denominator

Multiplying by $\frac{e^x}{e^x}$:

\[\frac{1}{1+3e^{-x}} = \frac{e^x}{e^x+3}\]

Therefore, the integral becomes:

\[\int_0^{\ln 5} \frac{\sqrt{e^x-1}}{1+3e^{-x}} \, dx = \int_0^{\ln 5} \frac{\sqrt{e^x-1} \cdot e^x}{e^x+3} \, dx\]

Step 2: Use substitution

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

Step 3: Change the limits of integration

When $x = 0$: $u = 0$. When $x = \ln 5$: $u = 5-1=4$.

Step 4: Express all terms in terms of $u$

• $\sqrt{e^x-1} = \sqrt{u}$
• $e^x + 3 = u + 4$
• $dx = \frac{du}{u+1}$

Step 5: Rewrite the integral

\[\int_0^{\ln 5} \frac{\sqrt{e^x-1} \cdot e^x}{e^x+3} \, dx = \int_0^4 \frac{\sqrt{u} \cdot (u+1)}{u+4} \cdot \frac{du}{u+1} = \int_0^4 \frac{\sqrt{u}}{u+4} \, du\]

Step 6: Use another substitution

Let $v = \sqrt{u}$, so $u = v^2$ and $du = 2v \, dv$. When $u = 0$: $v = 0$; when $u = 4$: $v = 2$.

Step 7: Rewrite in terms of $v$

\[\int_0^4 \frac{\sqrt{u}}{u+4} \, du = \int_0^2 \frac{2v^2}{v^2+4} \, dv\]

Step 8: Simplify using polynomial division

\[\frac{2v^2}{v^2+4} = 2 - \frac{8}{v^2+4}\]

Therefore:

\[\int_0^2 \frac{2v^2}{v^2+4} \, dv = \int_0^2 2 \, dv - \int_0^2 \frac{8}{v^2+4} \, dv\]

Step 9: Integrate term by term

\[\int_0^2 2 \, dv = 4\]

For the second integral, with $\int \frac{1}{v^2+a^2}\,dv=\frac1a\arctan\left(\frac va\right)+C$ and $a=2$:

\[\int_0^2 \frac{8}{v^2+4} \, dv = 4\left[\arctan\left(\frac{v}{2}\right)\right]_0^2 = 4\left[\frac{\pi}{4} - 0\right] = \pi\]

Step 10: Combine the results

\[\int_0^{\ln 5} \frac{\sqrt{e^x-1}}{1+3e^{-x}} \, dx = 4 - \pi\]

Numerical verification:

\[4 - \pi \approx 0.85841\]

This is a positive value, as expected for the area under a positive function.

Verification of bounds:

At $x=0$: the integrand is $\frac{\sqrt0}{1+3}=0$. At $x=\ln5$: the integrand is $\frac{\sqrt4}{1+3/5}=\frac{2}{8/5}=\frac54$. The function is positive and continuous on $[0,\ln5]$, so a positive result makes sense.

Final answer:

\[\boxed{\int_0^{\ln 5} \frac{\sqrt{e^x-1}}{1+3e^{-x}} \, dx = 4 - \pi \approx 0.8584}\]