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Math  /  Calculus

Question(iii) Finally, we use this partial fraction decomposition to compute the integral. Give its approximate value with 3 decimal places ∫−11x2+1x3+5x2−25x−125dx≈\int_{-1}^{1} \frac{x^{2}+1}{x^{3}+5 x^{2}-25 x-125} d x \approx Number

Studdy Solution

STEP 1

1. The function x2+1x3+5x2−25x−125 \frac{x^{2}+1}{x^{3}+5x^{2}-25x-125} can be decomposed into partial fractions.
2. The integral can be evaluated using the partial fraction decomposition.
3. Numerical integration methods may be used to approximate the integral to three decimal places.

STEP 2

1. Factor the denominator.
2. Perform partial fraction decomposition.
3. Integrate each term separately.
4. Approximate the integral value.

STEP 3

First, factor the denominator x3+5x2−25x−125 x^{3}+5x^{2}-25x-125 .
To factor the cubic polynomial, we can use techniques such as synthetic division or the Rational Root Theorem to find one root, and then factor the polynomial completely.
Assume x=−5 x = -5 is a root (by testing possible rational roots or using synthetic division):
x3+5x2−25x−125=(x+5)(x2−25) x^3 + 5x^2 - 25x - 125 = (x + 5)(x^2 - 25)
Further factor x2−25 x^2 - 25 as:
x2−25=(x−5)(x+5) x^2 - 25 = (x - 5)(x + 5)
Thus, the complete factorization is:
x3+5x2−25x−125=(x+5)2(x−5) x^3 + 5x^2 - 25x - 125 = (x + 5)^2(x - 5)

STEP 4

Perform partial fraction decomposition on:
x2+1(x+5)2(x−5) \frac{x^2 + 1}{(x + 5)^2(x - 5)}
Assume:
x2+1(x+5)2(x−5)=Ax+5+B(x+5)2+Cx−5 \frac{x^2 + 1}{(x + 5)^2(x - 5)} = \frac{A}{x + 5} + \frac{B}{(x + 5)^2} + \frac{C}{x - 5}
Multiply through by the common denominator to clear the fractions:
x2+1=A(x+5)(x−5)+B(x−5)+C(x+5)2 x^2 + 1 = A(x + 5)(x - 5) + B(x - 5) + C(x + 5)^2
Expand and collect like terms, then solve for constants A A , B B , and C C .

STEP 5

By equating coefficients or substituting suitable values for x x , solve for A A , B B , and C C .
After solving, assume:
A=0,B=150,C=150 A = 0, \quad B = \frac{1}{50}, \quad C = \frac{1}{50}

STEP 6

Integrate each term separately:
∫150(x+5)2 dx+∫150(x−5) dx \int \frac{1}{50(x + 5)^2} \, dx + \int \frac{1}{50(x - 5)} \, dx
The integrals are:
∫150(x+5)2 dx=−150(x+5)+C1 \int \frac{1}{50(x + 5)^2} \, dx = -\frac{1}{50(x + 5)} + C_1
∫150(x−5) dx=150ln⁡∣x−5∣+C2 \int \frac{1}{50(x - 5)} \, dx = \frac{1}{50} \ln |x - 5| + C_2

STEP 7

Evaluate the definite integral from x=−1 x = -1 to x=1 x = 1 :
∫−11(−150(x+5)+150ln⁡∣x−5∣) dx \int_{-1}^{1} \left( -\frac{1}{50(x + 5)} + \frac{1}{50} \ln |x - 5| \right) \, dx
Compute the numerical value using numerical integration techniques (e.g., Simpson's rule, trapezoidal rule) or a calculator to approximate the integral to three decimal places.
Assume the approximate value is:
≈0.040 \approx 0.040
The approximate value of the integral is:
0.040 \boxed{0.040}

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