Math Problem Statement

∫ x 2 + 1 x 2 − 5 x + 6 d x

Solution

It looks like you're asking for the integral:

x2+1x25x+6dx\int \frac{x^2 + 1}{x^2 - 5x + 6} \, dx

Let's solve this step by step.

Step 1: Factor the Denominator

First, let's factor the quadratic in the denominator:

x25x+6=(x2)(x3)x^2 - 5x + 6 = (x - 2)(x - 3)

Step 2: Simplify the Integrand

The integral becomes:

x2+1(x2)(x3)dx\int \frac{x^2 + 1}{(x - 2)(x - 3)} \, dx

Step 3: Divide the Numerator

To simplify, let's divide the numerator by the denominator:

We perform polynomial long division by dividing x2+1x^2 + 1 by x25x+6x^2 - 5x + 6.

x2+1x25x+6=1+(5x5)(x2)(x3)\frac{x^2 + 1}{x^2 - 5x + 6} = 1 + \frac{(5x - 5)}{(x - 2)(x - 3)}

Step 4: Partial Fraction Decomposition

Now, let's break down 5x5(x2)(x3)\frac{5x - 5}{(x - 2)(x - 3)} using partial fractions:

5x5(x2)(x3)=Ax2+Bx3\frac{5x - 5}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}

To find AA and BB:

5x5=A(x3)+B(x2)5x - 5 = A(x - 3) + B(x - 2)

Expanding and collecting terms:

5x5=(A+B)x(3A+2B)5x - 5 = (A + B)x - (3A + 2B)

Now, equate coefficients of like terms:

  1. Coefficient of xx: A+B=5A + B = 5
  2. Constant term: 3A2B=5-3A - 2B = -5

Step 5: Solve for AA and BB

Solve the system of equations:

A+B=5A + B = 5 3A2B=5-3A - 2B = -5

From the first equation, B=5AB = 5 - A. Substituting into the second equation:

3A2(5A)=5-3A - 2(5 - A) = -5 3A10+2A=5-3A - 10 + 2A = -5 A=5A=5-A = 5 \quad \Rightarrow \quad A = -5 B=5(5)=10B = 5 - (-5) = 10

Thus:

5x5(x2)(x3)=5x2+10x3\frac{5x - 5}{(x - 2)(x - 3)} = \frac{-5}{x - 2} + \frac{10}{x - 3}

Step 6: Integrate

Now, substitute back into the integral:

x2+1(x2)(x3)dx=(1+5x2+10x3)dx\int \frac{x^2 + 1}{(x - 2)(x - 3)} \, dx = \int \left(1 + \frac{-5}{x - 2} + \frac{10}{x - 3}\right) \, dx

This simplifies to:

x5lnx2+10lnx3+Cx - 5 \ln |x - 2| + 10 \ln |x - 3| + C

Where CC is the constant of integration.

Final Answer:

x2+1x25x+6dx=x5lnx2+10lnx3+C\int \frac{x^2 + 1}{x^2 - 5x + 6} \, dx = x - 5 \ln |x - 2| + 10 \ln |x - 3| + C

Let me know if you need further explanation or have any other questions!


  1. What are the basic steps in polynomial long division?
  2. How do you approach solving integrals that involve partial fractions?
  3. What is the significance of logarithms appearing in the final answer?
  4. Can partial fractions always be used when integrating rational functions?
  5. How does one determine the constants in partial fraction decomposition?

Tip: It's useful to check your integration result by differentiating it to see if you obtain the original integrand.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Partial Fraction Decomposition
Logarithmic Functions

Formulas

Partial fraction decomposition formula

Theorems

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Suitable Grade Level

College Level