Math Problem Statement

Integrate (x^3-2x^2+2x-7)/(x^2-2x-3) dx

Solution

To solve the integral

I=x32x2+2x7x22x3dx,I = \int \frac{x^3 - 2x^2 + 2x - 7}{x^2 - 2x - 3} \, dx,

we can proceed with polynomial long division, followed by partial fraction decomposition if necessary.

Step 1: Polynomial Long Division

Divide the numerator x32x2+2x7x^3 - 2x^2 + 2x - 7 by the denominator x22x3x^2 - 2x - 3.

  1. Divide x3x^3 by x2x^2, which gives xx.
  2. Multiply xx by the denominator x22x3x^2 - 2x - 3, resulting in x32x23xx^3 - 2x^2 - 3x.
  3. Subtract (x32x23x)(x^3 - 2x^2 - 3x) from the original numerator:

(x32x2+2x7)(x32x23x)=5x7.(x^3 - 2x^2 + 2x - 7) - (x^3 - 2x^2 - 3x) = 5x - 7.

Now, we need to divide the remainder 5x75x - 7 by x22x3x^2 - 2x - 3.

Since the degree of the remainder is lower than that of the denominator, we can now express the integral as:

I=(x+5x7x22x3)dx.I = \int (x + \frac{5x - 7}{x^2 - 2x - 3}) \, dx.

Step 2: Partial Fraction Decomposition

Factor the denominator x22x3x^2 - 2x - 3 as (x3)(x+1)(x - 3)(x + 1). So we want to decompose:

5x7(x3)(x+1)=Ax3+Bx+1.\frac{5x - 7}{(x - 3)(x + 1)} = \frac{A}{x - 3} + \frac{B}{x + 1}.

Multiply both sides by (x3)(x+1)(x - 3)(x + 1) to get:

5x7=A(x+1)+B(x3).5x - 7 = A(x + 1) + B(x - 3).

Expanding the right-hand side:

5x7=A(x+1)+B(x3)=Ax+A+Bx3B=(A+B)x+(A3B).5x - 7 = A(x + 1) + B(x - 3) = A x + A + B x - 3B = (A + B)x + (A - 3B).

Now, equate the coefficients of like terms:

  1. For xx-terms: A+B=5A + B = 5,
  2. For the constant terms: A3B=7A - 3B = -7.

Solving this system of equations:

  • From A+B=5A + B = 5, we have A=5BA = 5 - B.
  • Substituting into A3B=7A - 3B = -7:

(5B)3B=7    54B=7    4B=12    B=3.(5 - B) - 3B = -7 \implies 5 - 4B = -7 \implies -4B = -12 \implies B = 3.

  • Substituting B=3B = 3 into A+B=5A + B = 5, we get A=2A = 2.

Step 3: Rewrite the Integral

Now, we can express the integral as:

I=(x+2x3+3x+1)dx.I = \int \left( x + \frac{2}{x - 3} + \frac{3}{x + 1} \right) dx.

Step 4: Integrate Term by Term

Now integrate each term:

I=xdx+21x3dx+31x+1dx.I = \int x \, dx + 2 \int \frac{1}{x - 3} \, dx + 3 \int \frac{1}{x + 1} \, dx.

  • xdx=x22\int x \, dx = \frac{x^2}{2},
  • 21x3dx=2lnx32 \int \frac{1}{x - 3} \, dx = 2 \ln |x - 3|,
  • 31x+1dx=3lnx+13 \int \frac{1}{x + 1} \, dx = 3 \ln |x + 1|.

Thus, the final result is:

I=x22+2lnx3+3lnx+1+C,I = \frac{x^2}{2} + 2 \ln |x - 3| + 3 \ln |x + 1| + C,

where CC is the constant of integration.

Final Answer:

x32x2+2x7x22x3dx=x22+2lnx3+3lnx+1+C.\int \frac{x^3 - 2x^2 + 2x - 7}{x^2 - 2x - 3} \, dx = \frac{x^2}{2} + 2 \ln |x - 3| + 3 \ln |x + 1| + C.

Would you like more details or clarification on any step?

Here are 5 related questions for further practice:

  1. How would you integrate x34x+5x21dx\frac{x^3 - 4x + 5}{x^2 - 1} \, dx?
  2. What is the general approach to solve an integral involving rational functions?
  3. Can you explain how to use synthetic division in similar integration problems?
  4. How do you perform partial fraction decomposition for higher-degree polynomials?
  5. How would the integral change if the denominator had complex roots?

Tip: Always check if the degree of the numerator is greater than or equal to the denominator for polynomial long division.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Polynomial Long Division
Partial Fraction Decomposition

Formulas

Polynomial Long Division: Divide numerator by denominator when degree of numerator is higher
Partial Fraction Decomposition: A/(x - a) + B/(x - b) for factored denominators

Theorems

Fundamental Theorem of Calculus: Integration as the reverse process of differentiation
Partial Fraction Decomposition for Rational Functions

Suitable Grade Level

Undergraduate Calculus or AP Calculus