Math Problem Statement

5x2 + 9x − 1 /2x3 + 3x2 − 2x dx.

Solution

To solve the integral of the expression:

5x2+9x12x3+3x22xdx,\int \frac{5x^2 + 9x - 1}{2x^3 + 3x^2 - 2x} \, dx,

we can apply the method of partial fractions if the denominator can be factored, or we can attempt polynomial division if appropriate. Let's break it down:

  1. Factor the denominator if possible.
  2. Simplify the expression.
  3. Solve the integral using appropriate methods like polynomial division or partial fractions.

Step 1: Factor the denominator

The denominator is 2x3+3x22x2x^3 + 3x^2 - 2x. We can factor out an xx first:

2x3+3x22x=x(2x2+3x2).2x^3 + 3x^2 - 2x = x(2x^2 + 3x - 2).

Now, factor the quadratic 2x2+3x22x^2 + 3x - 2:

2x2+3x2=(2x1)(x+2).2x^2 + 3x - 2 = (2x - 1)(x + 2).

So, the denominator becomes:

2x3+3x22x=x(2x1)(x+2).2x^3 + 3x^2 - 2x = x(2x - 1)(x + 2).

Step 2: Rewrite the integral

Now, we rewrite the integral as:

5x2+9x1x(2x1)(x+2)dx.\int \frac{5x^2 + 9x - 1}{x(2x - 1)(x + 2)} \, dx.

We can attempt to decompose this expression into partial fractions:

5x2+9x1x(2x1)(x+2)=Ax+B2x1+Cx+2.\frac{5x^2 + 9x - 1}{x(2x - 1)(x + 2)} = \frac{A}{x} + \frac{B}{2x - 1} + \frac{C}{x + 2}.

We now need to determine the values of AA, BB, and CC.

Step 3: Determine coefficients for partial fractions

Multiplying both sides by x(2x1)(x+2)x(2x - 1)(x + 2), we get:

5x2+9x1=A(2x1)(x+2)+Bx(x+2)+Cx(2x1).5x^2 + 9x - 1 = A(2x - 1)(x + 2) + Bx(x + 2) + Cx(2x - 1).

Expanding both sides and equating coefficients of powers of xx will allow us to solve for AA, BB, and CC.

I'll proceed with this computation next.The partial fraction decomposition yields the following values for AA, BB, and CC:

A=12,B=195,C=110.A = \frac{1}{2}, \quad B = \frac{19}{5}, \quad C = \frac{1}{10}.

Thus, the integral becomes:

(1/2x+19/52x1+1/10x+2)dx.\int \left( \frac{1/2}{x} + \frac{19/5}{2x - 1} + \frac{1/10}{x + 2} \right) dx.

Step 4: Integrate each term

Now, we integrate each term separately:

  1. 1/2xdx=12lnx\int \frac{1/2}{x} \, dx = \frac{1}{2} \ln |x|
  2. 19/52x1dx=1910ln2x1\int \frac{19/5}{2x - 1} \, dx = \frac{19}{10} \ln |2x - 1|
  3. 1/10x+2dx=110lnx+2\int \frac{1/10}{x + 2} \, dx = \frac{1}{10} \ln |x + 2|

Final solution:

The final integral is:

12lnx+1910ln2x1+110lnx+2+C.\frac{1}{2} \ln |x| + \frac{19}{10} \ln |2x - 1| + \frac{1}{10} \ln |x + 2| + C.

Would you like any more details or have further questions?

Here are 5 related questions:

  1. How do we determine whether partial fractions are applicable to a given expression?
  2. Can polynomial division be used instead of partial fractions in some cases?
  3. How can integration by substitution help simplify complex integrals?
  4. What are some common techniques for integrating rational functions?
  5. How does factoring the denominator influence the solution method?

Tip: When dealing with integrals, always check if simplifying the expression first (e.g., factoring) can make the integration process easier.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integration
Partial Fractions
Polynomial Factorization

Formulas

Partial fraction decomposition: A/x + B/(2x-1) + C/(x+2)
Basic integration formulas: ∫(1/x) dx = ln|x|

Theorems

Fundamental Theorem of Calculus
Factorization of polynomials

Suitable Grade Level

Undergraduate Calculus