Math Problem Statement

2x2+2x18x36x2+9x\int\frac{2x^2+2x-18}{x^3-6x^2+9x}

Solution

To solve the integral

2x2+2x18x36x2+9xdx,\int \frac{2x^2 + 2x - 18}{x^3 - 6x^2 + 9x} \, dx,

we can simplify the integrand before performing the integration. Let's proceed step by step:

Step 1: Factor the denominator

We can factor the denominator x36x2+9xx^3 - 6x^2 + 9x as:

x36x2+9x=x(x26x+9)=x(x3)2.x^3 - 6x^2 + 9x = x(x^2 - 6x + 9) = x(x - 3)^2.

So the integral becomes:

2x2+2x18x(x3)2dx.\int \frac{2x^2 + 2x - 18}{x(x - 3)^2} \, dx.

Step 2: Use partial fraction decomposition

We now decompose the fraction:

2x2+2x18x(x3)2.\frac{2x^2 + 2x - 18}{x(x - 3)^2}.

Assume that:

2x2+2x18x(x3)2=Ax+Bx3+C(x3)2.\frac{2x^2 + 2x - 18}{x(x - 3)^2} = \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2}.

Multiply both sides by x(x3)2x(x - 3)^2 to get:

2x2+2x18=A(x3)2+Bx(x3)+Cx.2x^2 + 2x - 18 = A(x - 3)^2 + Bx(x - 3) + Cx.

Now, expand the right-hand side:

A(x3)2=A(x26x+9),A(x - 3)^2 = A(x^2 - 6x + 9), Bx(x3)=B(x23x),Bx(x - 3) = B(x^2 - 3x), Cx=Cx.Cx = Cx.

So, combining terms:

2x2+2x18=A(x26x+9)+B(x23x)+Cx.2x^2 + 2x - 18 = A(x^2 - 6x + 9) + B(x^2 - 3x) + Cx.

2x2+2x18=(A+B)x2+(6A3B+C)x+9A.2x^2 + 2x - 18 = (A + B)x^2 + (-6A - 3B + C)x + 9A.

Step 3: Set up a system of equations

Now, compare the coefficients of powers of xx on both sides:

  • Coefficient of x2x^2: A+B=2A + B = 2,
  • Coefficient of xx: 6A3B+C=2-6A - 3B + C = 2,
  • Constant term: 9A=189A = -18.

From the constant term equation, we have:

9A=18    A=2.9A = -18 \implies A = -2.

Substitute A=2A = -2 into the first equation:

2+B=2    B=4.-2 + B = 2 \implies B = 4.

Substitute A=2A = -2 and B=4B = 4 into the second equation:

6(2)3(4)+C=2    1212+C=2    C=2.-6(-2) - 3(4) + C = 2 \implies 12 - 12 + C = 2 \implies C = 2.

Step 4: Rewrite the integral

We now have the partial fractions:

2x2+2x18x(x3)2=2x+4x3+2(x3)2.\frac{2x^2 + 2x - 18}{x(x - 3)^2} = \frac{-2}{x} + \frac{4}{x - 3} + \frac{2}{(x - 3)^2}.

So, the integral becomes:

(2x+4x3+2(x3)2)dx.\int \left( \frac{-2}{x} + \frac{4}{x - 3} + \frac{2}{(x - 3)^2} \right) \, dx.

Step 5: Integrate each term

Now integrate term by term:

  1. 2xdx=2lnx\int \frac{-2}{x} \, dx = -2 \ln|x|,
  2. 4x3dx=4lnx3\int \frac{4}{x - 3} \, dx = 4 \ln|x - 3|,
  3. 2(x3)2dx=2x3\int \frac{2}{(x - 3)^2} \, dx = \frac{-2}{x - 3}.

Step 6: Final answer

Combining all the integrals, we get the final solution:

2x2+2x18x(x3)2dx=2lnx+4lnx32x3+C,\int \frac{2x^2 + 2x - 18}{x(x - 3)^2} \, dx = -2 \ln|x| + 4 \ln|x - 3| - \frac{2}{x - 3} + C,

where CC is the constant of integration.

Would you like further details or have any questions?


Here are 5 related questions:

  1. How do we approach partial fraction decomposition in general cases?
  2. What are other methods to simplify integrals involving rational functions?
  3. Can you explain why logarithmic functions appear in the solution?
  4. How do we handle improper integrals involving similar rational functions?
  5. How does the factorization of polynomials help in solving integrals?

Tip: Always check if you can simplify or factor the integrand before attempting more advanced techniques.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integral Calculus
Partial Fractions
Polynomial Factorization

Formulas

Partial Fraction Decomposition
Integration of 1/x: \( \int \frac{1}{x} dx = \ln |x| + C \)
Integration of 1/(x-a)^2: \( \int \frac{1}{(x-a)^2} dx = \frac{-1}{x-a} + C \)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12, University Calculus I