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The integral you've provided appears to be:

\[
\int \frac{dv}{v t + \sqrt{t F v^2}}
\]

This integral seems to involve variables \( v \), \( t \), and \( F \), and includes a square root term. To solve this, we can attempt a substitution or use standard integration techniques. Let me analyze the integral step by step:

### Step 1: Simplify the Denominator
The denominator is:

\[
v t + \sqrt{t F v^2}
\]

The square root term can be rewritten as:

\[
\sqrt{t F v^2} = v \sqrt{t F}
\]

Thus, the denominator becomes:

\[
v t + v \sqrt{t F} = v (t + \sqrt{t F})
\]

### Step 2: Rewrite the Integral
Now, the integral simplifies to:

\[
\int \frac{dv}{v (t + \sqrt{t F})}
\]

### Step 3: Factor Out Constants
Since \( t + \sqrt{t F} \) is independent of \( v \), we can factor it out of the integral:

\[
\frac{1}{t + \sqrt{t F}} \int \frac{dv}{v}
\]

### Step 4: Solve the Integral
The integral of \( \frac{1}{v} \) is a standard result:

\[
\int \frac{dv}{v} = \ln |v|
\]

Thus, the solution becomes:

\[
\frac{\ln |v|}{t + \sqrt{t F}} + C
\]

where \( C \) is the constant of integration.

### Final Answer:
\[
\int \frac{dv}{v t + \sqrt{t F v^2}} = \frac{\ln |v|}{t + \sqrt{t F}} + C
\]

Would you like more details or clarifications? Here are 5 related questions:

1. What if \( F \) or \( t \) were constants? How would this affect the solution?
2. How would this integral change if the square root term involved higher powers of \( v \)?
3. Could substitution methods provide a faster solution?
4. How does this solution relate to integrals involving logarithms and exponential functions?
5. What are some practical applications of integrals involving variables under square roots?

**Tip:** When working with integrals that contain square roots, always check if factoring or substitution can simplify the expression early on.

Explore the detailed solution for the integral ∫ (dv) / (v*t + √(t*F*v²)). This problem involves advanced calculus concepts such as integration and substitution methods, making it ideal for college-level calculus students.

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