definite integral 0 to 1 of 2/2x^2+3x+1 dx

We are tasked with solving the definite integral:

\[
\int_0^1 \frac{2}{2x^2 + 3x + 1} \, dx
\]

### Step 1: Simplify the Integral

First, factor the quadratic expression \( 2x^2 + 3x + 1 \):

\[
2x^2 + 3x + 1 = (2x + 1)(x + 1)
\]

Thus, the integral becomes:

\[
\int_0^1 \frac{2}{(2x + 1)(x + 1)} \, dx
\]

### Step 2: Partial Fraction Decomposition

To solve the integral, we decompose the fraction into partial fractions:

\[
\frac{2}{(2x + 1)(x + 1)} = \frac{A}{2x + 1} + \frac{B}{x + 1}
\]

Multiplying both sides by \((2x + 1)(x + 1)\) to eliminate the denominators:

\[
2 = A(x + 1) + B(2x + 1)
\]

Expanding both sides:

\[
2 = A(x + 1) + B(2x + 1) = A x + A + 2B x + B
\]

Group the terms by powers of \(x\):

\[
2 = (A + 2B)x + (A + B)
\]

Now, equate the coefficients of \(x\) and the constant term:

1. \( A + 2B = 0 \)
2. \( A + B = 2 \)

From the first equation: \( A = -2B \).

Substitute into the second equation:

\[
-2B + B = 2 \quad \Rightarrow \quad -B = 2 \quad \Rightarrow \quad B = -2
\]

Now, substitute \(B = -2\) into \(A = -2B\):

\[
A = -2(-2) = 4
\]

Thus, the partial fraction decomposition is:

\[
\frac{2}{(2x + 1)(x + 1)} = \frac{4}{2x + 1} - \frac{2}{x + 1}
\]

### Step 3: Integrate the Expression

Now, we can integrate:

\[
\int_0^1 \frac{4}{2x + 1} \, dx - \int_0^1 \frac{2}{x + 1} \, dx
\]

The integrals of these forms are straightforward logarithmic integrals:

\[
\int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln |ax + b|
\]

So, applying this rule:

1. \(\int \frac{4}{2x + 1} \, dx = 2 \ln |2x + 1|\)
2. \(\int \frac{2}{x + 1} \, dx = 2 \ln |x + 1|\)

Now, apply the limits:

\[
2 \left[ \ln(2x + 1) \right]_0^1 - 2 \left[ \ln(x + 1) \right]_0^1
\]

### Step 4: Compute the Values

1. For \( 2 \ln(2x + 1) \) from 0 to 1:
   \[
   2 \ln(3) - 2 \ln(1) = 2 \ln(3)
   \]

2. For \( 2 \ln(x + 1) \) from 0 to 1:
   \[
   2 \ln(2) - 2 \ln(1) = 2 \ln(2)
   \]

Thus, the result is:

\[
2 \ln(3) - 2 \ln(2) = 2 \ln\left(\frac{3}{2}\right)
\]

### Final Answer:

\[
\boxed{2 \ln\left(\frac{3}{2}\right)}
\]

Would you like further details or clarification?

Here are 5 related questions:
1. How do you perform partial fraction decomposition for different types of quadratics?
2. Can you apply logarithmic rules to simplify integrals?
3. How do integrals change when bounds are negative or involve infinity?
4. How can you integrate functions involving higher powers of \(x\)?
5. What is the role of substitution in definite integrals?

**Tip:** Always check if a quadratic expression can be factored before attempting partial fractions.

Learn how to solve the definite integral of 2/(2x^2 + 3x + 1) from 0 to 1 using partial fraction decomposition and logarithmic integration. This solution involves key calculus concepts, including the use of logarithmic integrals and partial fractions. Suitable for college-level calculus students.

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