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Let's evaluate the integral given:

\[
\int_1^x \frac{2}{t^2} \, dt
\]

To solve this definite integral, we first need to determine the indefinite integral of the integrand \(\frac{2}{t^2}\).

### Step 1: Find the Indefinite Integral
The integrand is \(\frac{2}{t^2}\). This can be rewritten as:

\[
2 \cdot t^{-2}
\]

Now integrate:

\[
\int 2t^{-2} \, dt = 2 \int t^{-2} \, dt
\]

Using the power rule for integration:

\[
\int t^n \, dt = \frac{t^{n+1}}{n+1} \quad \text{(for } n \neq -1\text{)}
\]

Here, \(n = -2\):

\[
\int t^{-2} \, dt = \frac{t^{-1}}{-1} = -t^{-1} = -\frac{1}{t}
\]

So:

\[
\int 2t^{-2} \, dt = -\frac{2}{t}
\]

### Step 2: Evaluate the Definite Integral
Now we evaluate the definite integral from 1 to \(x\):

\[
\int_1^x \frac{2}{t^2} \, dt = \left[ -\frac{2}{t} \right]_1^x
\]

First, substitute the limits:

\[
\left[ -\frac{2}{t} \right]_1^x = -\frac{2}{x} - \left( -\frac{2}{1} \right)
\]

\[
= -\frac{2}{x} + 2
\]

\[
= 2 - \frac{2}{x}
\]

### Step 3: Simplify the Expression
We can rewrite:

\[
2 - \frac{2}{x} = \frac{2x - 2}{x} = \frac{2(x - 1)}{x}
\]

### Final Answer
The correct answer is:

\[
\boxed{\text{B. } \frac{2(x-1)}{x}}
\]

Would you like more details on any step, or do you have further questions?

### Related Questions
1. How do you integrate functions with negative exponents?
2. What are the properties of definite integrals that can help solve this problem?
3. Can the Fundamental Theorem of Calculus be used to find definite integrals?
4. How do you solve integrals involving fractional powers?
5. How can we verify definite integrals graphically?

### Tip
Always simplify your result as much as possible to match the answer choices, especially in multiple-choice questions.

This detailed solution covers how to evaluate the definite integral of 2/t² from 1 to x. It uses key calculus concepts such as the power rule for integration and the Fundamental Theorem of Calculus. Suitable for high school seniors or early university students learning integral calculus.

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