To evaluate the integral:

\[
\int \frac{3x + 2}{\sqrt{1 - 2x}} \, dx
\]

we can proceed by using substitution.

### Step 1: Substitution
Let:
\[
u = 1 - 2x
\]
Then, differentiate \( u \) with respect to \( x \):
\[
du = -2 \, dx \quad \Rightarrow \quad dx = -\frac{du}{2}
\]

Now, we need to express \( 3x + 2 \) in terms of \( u \). From the substitution \( u = 1 - 2x \), we can solve for \( x \):
\[
x = \frac{1 - u}{2}
\]
Substitute this into \( 3x + 2 \):
\[
3x + 2 = 3\left(\frac{1 - u}{2}\right) + 2 = \frac{3(1 - u)}{2} + 2 = \frac{3 - 3u}{2} + \frac{4}{2} = \frac{7 - 3u}{2}
\]

### Step 2: Substituting into the integral
Substitute the expressions for \( 3x + 2 \) and \( dx \) into the integral:
\[
\int \frac{3x + 2}{\sqrt{1 - 2x}} \, dx = \int \frac{\frac{7 - 3u}{2}}{\sqrt{u}} \cdot -\frac{du}{2}
\]

This simplifies to:
\[
-\frac{1}{4} \int \frac{7 - 3u}{\sqrt{u}} \, du
\]
Now, split the integral into two parts:
\[
-\frac{1}{4} \left( 7 \int \frac{1}{\sqrt{u}} \, du - 3 \int \frac{u}{\sqrt{u}} \, du \right)
\]

### Step 3: Solving the integrals
Both integrals are straightforward:
1. \(\int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u}\)
2. \(\int \frac{u}{\sqrt{u}} \, du = \int \sqrt{u} \, du = \frac{2u^{3/2}}{3}\)

Substitute these into the expression:
\[
-\frac{1}{4} \left( 7 \cdot 2\sqrt{u} - 3 \cdot \frac{2u^{3/2}}{3} \right)
= -\frac{1}{4} \left( 14\sqrt{u} - 2u^{3/2} \right)
= -\frac{7\sqrt{u}}{2} + \frac{u^{3/2}}{2}
\]

### Step 4: Back-substitute \( u = 1 - 2x \)
Now, substitute \( u = 1 - 2x \) back into the expression:
\[
-\frac{7\sqrt{1 - 2x}}{2} + \frac{(1 - 2x)^{3/2}}{2} + C
\]

Thus, the final result is:
\[
\boxed{-\frac{7\sqrt{1 - 2x}}{2} + \frac{(1 - 2x)^{3/2}}{2} + C}
\]

Let me know if you need further clarification or if you would like to explore any specific step in more detail.

Here are 5 related questions to deepen your understanding:
1. How can we apply substitution in integrals involving square roots?
2. What are the common substitution methods for integrals with rational functions?
3. How do we differentiate between a simple substitution and a trigonometric substitution?
4. Can we approach this problem using a different method, such as integration by parts?
5. What are the techniques to handle integrals that involve expressions like \( \sqrt{1 - x^2} \)?

**Tip:** When using substitution, always remember to adjust the limits of integration if you are solving a definite integral.

Learn how to evaluate the integral of (3x + 2)/√(1 - 2x) using substitution. This problem involves key calculus concepts like substitution and square root integrals, with a step-by-step guide to solving. Suitable for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation