https://file.aiquickdraw.com/mathbot/file/1730508316943nivcdloh.jpg

The integral in your image is:

\[
\int \frac{1}{\sqrt{2x - 1}} \, dx
\]

To solve this, we can use a substitution method. Here’s the step-by-step solution:

### Step 1: Substitution
Let:
\[
u = 2x - 1
\]

Then, differentiate \( u \) with respect to \( x \):
\[
\frac{du}{dx} = 2 \Rightarrow dx = \frac{du}{2}
\]

### Step 2: Substitute in Terms of \( u \)
Substitute \( u \) and \( dx \) in the integral:
\[
\int \frac{1}{\sqrt{2x - 1}} \, dx = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-\frac{1}{2}} \, du
\]

### Step 3: Integrate with Respect to \( u \)
Now, integrate \( u^{-\frac{1}{2}} \):
\[
\frac{1}{2} \int u^{-\frac{1}{2}} \, du = \frac{1}{2} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} = u^{\frac{1}{2}} = \sqrt{u}
\]

### Step 4: Substitute Back in Terms of \( x \)
Replace \( u \) with \( 2x - 1 \):
\[
\sqrt{u} = \sqrt{2x - 1}
\]

Thus, the final answer is:
\[
\int \frac{1}{\sqrt{2x - 1}} \, dx = \sqrt{2x - 1} + C
\]

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

---

Do you have any questions, or would you like more details on any step?

### Related Questions
1. How do you solve integrals that involve square roots?
2. What other types of substitutions work well for integrals?
3. How do we handle integrals involving higher powers of \( x \) in the denominator?
4. What are common techniques for integrating rational functions?
5. How does substitution help simplify complex integrals?

### Tip
For integrals involving expressions like \( \sqrt{ax + b} \), substitution with \( u = ax + b \) is a useful method to simplify the expression.

Evaluate the integral \( \int \frac{1}{\sqrt{2x - 1}} \, dx \)

Learn how to evaluate the integral \( \int \frac{1}{\sqrt{2x - 1}} \, dx \) using the substitution method. This calculus problem covers key integration techniques, ideal for students in Grades 11-12.

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