To solve the integral of the given function, we will first identify the expression to integrate. The expression is:

$\int \sqrt{\frac{mg}{b}} \times \frac{-1 - e^{2\sqrt{\frac{gb}{m}}t}}{1 - e^{2\sqrt{\frac{gb}{m}}t}} \, dt$

Step 1: Substituting constants

For simplicity, let's substitute constants to simplify the integral:

• Let $A = \sqrt{\frac{mg}{b}}$
• Let $\alpha = \sqrt{\frac{gb}{m}}$

Now the integral becomes:

$A \int \frac{-1 - e^{2\alpha t}}{1 - e^{2\alpha t}} \, dt$

Step 2: Simplifying the integrand

The expression we are integrating simplifies to:

$A \int \left( -1 - \frac{2 e^{\alpha t}}{1 - e^{2\alpha t}} \right) \, dt$

We now separate the integral into two parts:

$A \left( \int -1 \, dt - 2 \int \frac{e^{\alpha t}}{1 - e^{2\alpha t}} \, dt \right)$

Step 3: Solving the first integral

The first part is straightforward:

$\int -1 \, dt = -t$

Step 4: Solving the second integral

For the second integral, let's use substitution. Let $u = e^{\alpha t}$, so that $du = \alpha e^{\alpha t} dt$. Thus:

$\int \frac{e^{\alpha t}}{1 - e^{2\alpha t}} \, dt = \frac{1}{\alpha} \int \frac{du}{1 - u^2}$

This is a standard integral of the form:

$\int \frac{du}{1 - u^2} = \frac{1}{2} \ln \left| \frac{1+u}{1-u} \right|$

Substituting back $u = e^{\alpha t}$, we get:

$\frac{1}{\alpha} \times \frac{1}{2} \ln \left| \frac{1+e^{\alpha t}}{1-e^{\alpha t}} \right|$

Step 5: Combining the results

Now, combining the results from both integrals, we get:

$A \left( -t - \frac{1}{\alpha} \ln \left| \frac{1+e^{\alpha t}}{1-e^{\alpha t}} \right| \right)$

Substituting $A = \sqrt{\frac{mg}{b}}$ and $\alpha = \sqrt{\frac{gb}{m}}$, the final answer is:

$\sqrt{\frac{mg}{b}} \left( -t - \frac{1}{\sqrt{\frac{gb}{m}}} \ln \left| \frac{1+e^{\sqrt{\frac{gb}{m}} t}}{1-e^{\sqrt{\frac{gb}{m}} t}} \right| \right)$

Let me know if you'd like further clarification or details.

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Related Questions:

1. How does substitution work in integral solving?
2. Can you explain integrals involving logarithmic functions?
3. How do we interpret integrals involving exponentials?
4. What are the common techniques to solve complex integrals?
5. How does a constant multiplier affect the outcome of an integral?

Tip: When encountering complex integrals, breaking them into simpler parts makes it easier to approach the solution.