The image shows the following equation involving two integrals:

$\int \frac{1 - v}{v^2} dv = \int \frac{1}{x} dx$

Let's solve both integrals.

1. Left-hand Side: $\int \frac{1 - v}{v^2} dv$

We can split this integral into two separate terms:

$\int \frac{1 - v}{v^2} dv = \int \frac{1}{v^2} dv - \int \frac{v}{v^2} dv$

Solving $\int \frac{1}{v^2} dv$:

$\int v^{-2} dv = \frac{v^{-1}}{-1} = -\frac{1}{v}$

Solving $\int \frac{v}{v^2} dv$:

$\int v^{-1} dv = \ln|v|$

Thus, the left-hand side becomes: $-\frac{1}{v} - \ln|v|$

2. Right-hand Side: $\int \frac{1}{x} dx$

This is a standard integral: $\int \frac{1}{x} dx = \ln|x|$

Final Equation:

Thus, the equation becomes: $-\frac{1}{v} - \ln|v| = \ln|x| + C$ where $C$ is the constant of integration.

If you'd like further clarification or additional steps, let me know!

Follow-up Questions:

1. Can you verify the conditions for when $v$ and $x$ are positive or negative?
2. How would the solution change if we integrate with respect to definite limits?
3. Can you substitute specific values for $v$ and $x$ to check the solution?
4. What is the physical interpretation of this equation, if it comes from a real-world context?
5. How would you find $v$ as a function of $x$ from the resulting equation?

Tip:

Always remember to include the constant of integration when solving indefinite integrals.