Math  /  Calculus

Question3x22x4x3x24x+4dx=\int \frac{3 x^{2}-2 x-4}{x^{3}-x^{2}-4 x+4} d x=

Studdy Solution

STEP 1

1. The integral is a rational function, which suggests the use of partial fraction decomposition.
2. The degree of the numerator is less than the degree of the denominator, which is appropriate for partial fraction decomposition.
3. The denominator can be factored to facilitate the decomposition.

STEP 2

1. Factor the denominator.
2. Set up the partial fraction decomposition.
3. Solve for the coefficients in the decomposition.
4. Integrate each term separately.

STEP 3

First, factor the denominator x3x24x+4 x^3 - x^2 - 4x + 4 . We will use synthetic division or polynomial factoring techniques to find the roots.
Let's try factoring by grouping:
x3x24x+4=(x3x2)(4x4) x^3 - x^2 - 4x + 4 = (x^3 - x^2) - (4x - 4) =x2(x1)4(x1) = x^2(x - 1) - 4(x - 1) =(x24)(x1) = (x^2 - 4)(x - 1)
Notice that x24 x^2 - 4 is a difference of squares:
x24=(x2)(x+2) x^2 - 4 = (x - 2)(x + 2)
Thus, the complete factorization of the denominator is:
(x2)(x+2)(x1) (x - 2)(x + 2)(x - 1)

STEP 4

Set up the partial fraction decomposition for the integrand:
3x22x4(x2)(x+2)(x1)=Ax2+Bx+2+Cx1 \frac{3x^2 - 2x - 4}{(x - 2)(x + 2)(x - 1)} = \frac{A}{x - 2} + \frac{B}{x + 2} + \frac{C}{x - 1}

STEP 5

Multiply through by the common denominator (x2)(x+2)(x1)(x - 2)(x + 2)(x - 1) to clear the fractions:
3x22x4=A(x+2)(x1)+B(x2)(x1)+C(x2)(x+2) 3x^2 - 2x - 4 = A(x + 2)(x - 1) + B(x - 2)(x - 1) + C(x - 2)(x + 2)
Expand and collect like terms to solve for A A , B B , and C C .

STEP 6

Substitute convenient values for x x to solve for the coefficients:
1. Let x=2 x = 2 :
3(2)22(2)4=A(2+2)(21) 3(2)^2 - 2(2) - 4 = A(2 + 2)(2 - 1) 1244=4A 12 - 4 - 4 = 4A 4=4A 4 = 4A A=1 A = 1
2. Let x=2 x = -2 :
3(2)22(2)4=B(22)(21) 3(-2)^2 - 2(-2) - 4 = B(-2 - 2)(-2 - 1) 12+44=12B 12 + 4 - 4 = 12B 12=12B 12 = 12B B=1 B = 1
3. Let x=1 x = 1 :
3(1)22(1)4=C(12)(1+2) 3(1)^2 - 2(1) - 4 = C(1 - 2)(1 + 2) 324=3C 3 - 2 - 4 = -3C 3=3C -3 = -3C C=1 C = 1

STEP 7

Now integrate each term separately:
(1x2+1x+2+1x1)dx \int \left( \frac{1}{x - 2} + \frac{1}{x + 2} + \frac{1}{x - 1} \right) dx
=1x2dx+1x+2dx+1x1dx = \int \frac{1}{x - 2} dx + \int \frac{1}{x + 2} dx + \int \frac{1}{x - 1} dx
Each integral is a natural logarithm:
=lnx2+lnx+2+lnx1+C = \ln|x - 2| + \ln|x + 2| + \ln|x - 1| + C
The solution to the integral is:
lnx2+lnx+2+lnx1+C \ln|x - 2| + \ln|x + 2| + \ln|x - 1| + C

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