Math Problem Statement

\int_{0}^{1} f(x) , dx f(x) = \frac{x+2}{(x+3)(x-2)}

Solution

We are tasked with evaluating the integral:

I=01x+2(x+3)(x2)dxI = \int_{0}^{1} \frac{x+2}{(x+3)(x-2)} \, dx

Step 1: Partial Fraction Decomposition

First, let's perform a partial fraction decomposition of the integrand:

x+2(x+3)(x2)=Ax+3+Bx2\frac{x+2}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}

To find the constants AA and BB, we multiply both sides by the denominator (x+3)(x2)(x+3)(x-2):

x+2=A(x2)+B(x+3)x+2 = A(x-2) + B(x+3)

Now expand the right-hand side:

x+2=A(x2)+B(x+3)=A(x)2A+B(x)+3B=(A+B)x+(2A+3B)x + 2 = A(x-2) + B(x+3) = A(x) - 2A + B(x) + 3B = (A + B)x + (-2A + 3B)

Equating the coefficients of like powers of xx from both sides:

  • For xx: A+B=1A + B = 1
  • For the constant term: 2A+3B=2-2A + 3B = 2

We now solve this system of equations:

  1. A+B=1A + B = 1
  2. 2A+3B=2-2A + 3B = 2

Step 2: Solving for AA and BB

From the first equation, solve for AA:

A=1BA = 1 - B

Substitute this into the second equation:

2(1B)+3B=2-2(1 - B) + 3B = 2 2+2B+3B=2-2 + 2B + 3B = 2 5B=45B = 4 B=45B = \frac{4}{5}

Now substitute B=45B = \frac{4}{5} into A+B=1A + B = 1:

A+45=1A + \frac{4}{5} = 1 A=145=15A = 1 - \frac{4}{5} = \frac{1}{5}

Step 3: Rewriting the Integral

Thus, we have the decomposition:

x+2(x+3)(x2)=1/5x+3+4/5x2\frac{x+2}{(x+3)(x-2)} = \frac{1/5}{x+3} + \frac{4/5}{x-2}

We can now rewrite the integral as:

I=01(1/5x+3+4/5x2)dxI = \int_{0}^{1} \left( \frac{1/5}{x+3} + \frac{4/5}{x-2} \right) dx

Step 4: Integrating Each Term

We integrate each term separately:

011x+3dx=lnx+301=ln(4)ln(3)\int_{0}^{1} \frac{1}{x+3} \, dx = \ln|x+3| \Big|_0^1 = \ln(4) - \ln(3) 011x2dx=lnx201=ln(1)ln(2)=ln(2)\int_{0}^{1} \frac{1}{x-2} \, dx = \ln|x-2| \Big|_0^1 = \ln(1) - \ln(2) = -\ln(2)

Thus, the integral becomes:

I=15(ln(4)ln(3))+45(ln(2))I = \frac{1}{5} \left( \ln(4) - \ln(3) \right) + \frac{4}{5} (-\ln(2))

Simplifying:

I=15ln(43)45ln(2)I = \frac{1}{5} \ln \left( \frac{4}{3} \right) - \frac{4}{5} \ln(2)

Final Answer:

The value of the integral is:

I=15ln(43)45ln(2)I = \frac{1}{5} \ln \left( \frac{4}{3} \right) - \frac{4}{5} \ln(2)

Would you like further details on any of the steps or have any questions? Here are some related questions to explore:

  1. What is the general method of solving partial fraction decompositions?
  2. How do you handle logarithmic integrals like this one in more complex cases?
  3. Can you compute the numerical value of this integral using logarithmic properties?
  4. How would this integral behave if the limits were changed, say from [0,2][0, 2]?
  5. What happens if the denominator has repeated factors in partial fraction decomposition?

Tip: When solving partial fractions, always ensure to match the degree of the numerators and denominators.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Partial Fraction Decomposition
Logarithmic Integration

Formulas

Partial Fraction Decomposition: \( \frac{x+2}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2} \)
Definite Integral of \( \frac{1}{x+a} \): \( \int \frac{1}{x+a} \, dx = \ln|x+a| + C \)

Theorems

Fundamental Theorem of Calculus
Properties of Logarithms

Suitable Grade Level

Undergraduate Level (Calculus I or II)