Let's solve Exercise 20 step by step. The given expression is:

$\frac{2x^4 + 9x^2 + x - 4}{x^3 + 4x}.$

Step 1: Factor the denominator

The denominator $x^3 + 4x$ can be factored as: $x^3 + 4x = x(x^2 + 4).$

Step 2: Divide numerator by denominator

The degree of the numerator $2x^4 + 9x^2 + x - 4$ is higher than the denominator $x^3 + 4x$. Therefore, we perform long division.

Long Division:

1. Divide the leading term of the numerator $2x^4$ by the leading term of the denominator $x^3$: $\frac{2x^4}{x^3} = 2x.$ Multiply $2x$ by $x^3 + 4x$ to get: $2x(x^3 + 4x) = 2x^4 + 8x^2.$ Subtract $2x^4 + 8x^2$ from the numerator: $(2x^4 + 9x^2 + x - 4) - (2x^4 + 8x^2) = x^2 + x - 4.$

2. The remainder is now $x^2 + x - 4$. Next, divide $x^2$ by $x^3$: this is not possible because the degree of the remainder is less than the denominator. Thus, the remainder becomes: $\frac{x^2 + x - 4}{x(x^2 + 4)}.$

Step 3: Partial fraction decomposition

We decompose: $\frac{x^2 + x - 4}{x(x^2 + 4)}.$

Write it as: $\frac{x^2 + x - 4}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}.$

Multiply through by $x(x^2 + 4)$ to get: $x^2 + x - 4 = A(x^2 + 4) + (Bx + C)(x).$

Expand: $x^2 + x - 4 = A(x^2) + 4A + Bx^2 + Cx.$

Combine like terms: $x^2 + x - 4 = (A + B)x^2 + Cx + 4A.$

Now, equate coefficients on both sides:

1. Coefficient of $x^2$: $A + B = 1$,
2. Coefficient of $x$: $C = 1$,
3. Constant term: $4A = -4$.

Solve for $A, B,$ and $C$:

• From $4A = -4$: $A = -1$,
• From $A + B = 1$: $-1 + B = 1$ → $B = 2$,
• From $C = 1$: $C = 1$.

Thus: $\frac{x^2 + x - 4}{x(x^2 + 4)} = \frac{-1}{x} + \frac{2x + 1}{x^2 + 4}.$

Step 4: Final result

Combine everything: $\frac{2x^4 + 9x^2 + x - 4}{x^3 + 4x} = 2x + \frac{-1}{x} + \frac{2x + 1}{x^2 + 4}.$

Step 5: Antiderivative (Optional)

If the problem asks for the antiderivative, integrate each term: $\int \left( 2x + \frac{-1}{x} + \frac{2x + 1}{x^2 + 4} \right) dx.$

1. $\int 2x \, dx = x^2$,
2. $\int \frac{-1}{x} \, dx = -\ln|x|$,
3. Split $\frac{2x + 1}{x^2 + 4}$:
   • $\int \frac{2x}{x^2 + 4} \, dx = \ln(x^2 + 4)$,
   • $\int \frac{1}{x^2 + 4} \, dx = \frac{1}{2} \arctan\left(\frac{x}{2}\right)$.

Thus: $\int \frac{2x^4 + 9x^2 + x - 4}{x^3 + 4x} dx = x^2 - \ln|x| + \ln(x^2 + 4) + \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C.$

Would you like a detailed explanation of any step, or should I solve another problem?

5 Related Questions:

1. How do you perform long division with polynomials?
2. What are the steps to factor a polynomial?
3. How do you solve partial fraction decomposition for higher-order denominators?
4. What are the key rules for integrating rational functions?
5. How can you identify when long division is needed in a fraction?

Tip:

When solving partial fractions, always check if long division is required first. This simplifies the process significantly!