Math Problem Statement

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

Solution

We are tasked with solving the definite integral:

0122x2+3x+1dx\int_0^1 \frac{2}{2x^2 + 3x + 1} \, dx

Step 1: Simplify the Integral

First, factor the quadratic expression 2x2+3x+12x^2 + 3x + 1:

2x2+3x+1=(2x+1)(x+1)2x^2 + 3x + 1 = (2x + 1)(x + 1)

Thus, the integral becomes:

012(2x+1)(x+1)dx\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:

2(2x+1)(x+1)=A2x+1+Bx+1\frac{2}{(2x + 1)(x + 1)} = \frac{A}{2x + 1} + \frac{B}{x + 1}

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

2=A(x+1)+B(2x+1)2 = A(x + 1) + B(2x + 1)

Expanding both sides:

2=A(x+1)+B(2x+1)=Ax+A+2Bx+B2 = A(x + 1) + B(2x + 1) = A x + A + 2B x + B

Group the terms by powers of xx:

2=(A+2B)x+(A+B)2 = (A + 2B)x + (A + B)

Now, equate the coefficients of xx and the constant term:

  1. A+2B=0A + 2B = 0
  2. A+B=2A + B = 2

From the first equation: A=2BA = -2B.

Substitute into the second equation:

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

Now, substitute B=2B = -2 into A=2BA = -2B:

A=2(2)=4A = -2(-2) = 4

Thus, the partial fraction decomposition is:

2(2x+1)(x+1)=42x+12x+1\frac{2}{(2x + 1)(x + 1)} = \frac{4}{2x + 1} - \frac{2}{x + 1}

Step 3: Integrate the Expression

Now, we can integrate:

0142x+1dx012x+1dx\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:

1ax+bdx=1alnax+b\int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln |ax + b|

So, applying this rule:

  1. 42x+1dx=2ln2x+1\int \frac{4}{2x + 1} \, dx = 2 \ln |2x + 1|
  2. 2x+1dx=2lnx+1\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 2ln(2x+1)2 \ln(2x + 1) from 0 to 1: 2ln(3)2ln(1)=2ln(3)2 \ln(3) - 2 \ln(1) = 2 \ln(3)

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

Thus, the result is:

2ln(3)2ln(2)=2ln(32)2 \ln(3) - 2 \ln(2) = 2 \ln\left(\frac{3}{2}\right)

Final Answer:

2ln(32)\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 xx?
  5. What is the role of substitution in definite integrals?

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

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Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integral
Partial Fraction Decomposition
Logarithmic Integration

Formulas

Integral of 1/(ax + b) = (1/a) * ln|ax + b|
Partial Fraction Decomposition: A/(2x+1) + B/(x+1)

Theorems

Fundamental Theorem of Calculus
Logarithmic Integration
Partial Fraction Decomposition

Suitable Grade Level

College-Level Calculus