Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
Let's tackle the problem step-by-step.
### Step 1: Partial Fraction Decomposition
We need to decompose the integrand into partial fractions.
The integrand is:
[tex]\[ \frac{x^2 - x + 6}{x^3 + 3x} \][/tex]
First, we factor the denominator:
[tex]\[ x^3 + 3x = x(x^2 + 3) \][/tex]
So, we can express the integrand as:
[tex]\[ \frac{x^2 - x + 6}{x(x^2 + 3)} \][/tex]
We assume a partial fraction decomposition of the form:
[tex]\[ \frac{x^2 - x + 6}{x(x^2 + 3)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 3} \][/tex]
Next, we multiply both sides by the denominator [tex]\( x(x^2 + 3) \)[/tex] to clear the fractions:
[tex]\[ x^2 - x + 6 = A(x^2 + 3) + (Bx + C)x \][/tex]
Now we expand and combine like terms:
[tex]\[ x^2 - x + 6 = Ax^2 + 3A + Bx^2 + Cx \][/tex]
[tex]\[ x^2 - x + 6 = (A + B)x^2 + Cx + 3A \][/tex]
We equate the coefficients of the corresponding powers of [tex]\(x\)[/tex] on both sides:
1. For [tex]\(x^2\)[/tex]: [tex]\( A + B = 1 \)[/tex]
2. For [tex]\(x\)[/tex]: [tex]\( C = -1 \)[/tex]
3. For the constant term: [tex]\( 3A = 6 \)[/tex]
From equation (3):
[tex]\[ 3A = 6 \Rightarrow A = 2 \][/tex]
Substitute [tex]\( A = 2 \)[/tex] into equation (1):
[tex]\[ 2 + B = 1 \Rightarrow B = -1 \][/tex]
So, the partial fraction decomposition is:
[tex]\[ \frac{2}{x} + \frac{(-1)x - 1}{x^2 + 3} \][/tex]
[tex]\[ = \frac{2}{x} - \frac{x + 1}{x^2 + 3} \][/tex]
### Step 2: Integrate the Partial Fractions
Now, we integrate each term separately:
[tex]\[ \int \left( \frac{2}{x} - \frac{x + 1}{x^2 + 3} \right) dx \][/tex]
First, integrate [tex]\( \frac{2}{x} \)[/tex]:
[tex]\[ \int \frac{2}{x} dx = 2 \ln |x| \][/tex]
Next, integrate [tex]\( \frac{x + 1}{x^2 + 3} \)[/tex]. Break it into two integrals:
[tex]\[ \int \frac{x + 1}{x^2 + 3} dx = \int \frac{x}{x^2 + 3} dx + \int \frac{1}{x^2 + 3} dx \][/tex]
For [tex]\( \int \frac{x}{x^2 + 3} dx \)[/tex]:
Use the substitution [tex]\( u = x^2 + 3 \)[/tex], then [tex]\( du = 2x \, dx \)[/tex]:
[tex]\[ \int \frac{x}{x^2 + 3} dx = \frac{1}{2} \int \frac{2x}{x^2 + 3} dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| = \frac{1}{2} \ln|x^2 + 3| \][/tex]
For [tex]\( \int \frac{1}{x^2 + 3} dx \)[/tex]:
Recall the integral of [tex]\( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan \left( \frac{x}{a} \right) \)[/tex]:
[tex]\[ \int \frac{1}{x^2 + 3} dx = \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
Combining these integrals, we get:
[tex]\[ \int \frac{x + 1}{x^2 + 3} dx = \frac{1}{2} \ln|x^2 + 3| + \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
Thus, the integral becomes:
[tex]\[ \int \left( \frac{2}{x} - \frac{x + 1}{x^2 + 3} \right) dx \][/tex]
[tex]\[ = 2 \ln |x| - \left( \frac{1}{2} \ln |x^2 + 3| + \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \right) \][/tex]
[tex]\[ = 2 \ln |x| - \frac{1}{2} \ln |x^2 + 3| - \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
### Final Answer
Thus, the partial fraction decomposition of:
[tex]\[ \frac{x^2 - x + 6}{x^3 + 3x} \][/tex]
is:
[tex]\[ \frac{2}{x} - \frac{x + 1}{x^2 + 3} \][/tex]
And the integral is:
[tex]\[ \int \frac{x^2 - x + 6}{x^3 + 3x} dx = 2 \ln |x| - \frac{1}{2} \ln |x^2 + 3| - \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) + C \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
### Step 1: Partial Fraction Decomposition
We need to decompose the integrand into partial fractions.
The integrand is:
[tex]\[ \frac{x^2 - x + 6}{x^3 + 3x} \][/tex]
First, we factor the denominator:
[tex]\[ x^3 + 3x = x(x^2 + 3) \][/tex]
So, we can express the integrand as:
[tex]\[ \frac{x^2 - x + 6}{x(x^2 + 3)} \][/tex]
We assume a partial fraction decomposition of the form:
[tex]\[ \frac{x^2 - x + 6}{x(x^2 + 3)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 3} \][/tex]
Next, we multiply both sides by the denominator [tex]\( x(x^2 + 3) \)[/tex] to clear the fractions:
[tex]\[ x^2 - x + 6 = A(x^2 + 3) + (Bx + C)x \][/tex]
Now we expand and combine like terms:
[tex]\[ x^2 - x + 6 = Ax^2 + 3A + Bx^2 + Cx \][/tex]
[tex]\[ x^2 - x + 6 = (A + B)x^2 + Cx + 3A \][/tex]
We equate the coefficients of the corresponding powers of [tex]\(x\)[/tex] on both sides:
1. For [tex]\(x^2\)[/tex]: [tex]\( A + B = 1 \)[/tex]
2. For [tex]\(x\)[/tex]: [tex]\( C = -1 \)[/tex]
3. For the constant term: [tex]\( 3A = 6 \)[/tex]
From equation (3):
[tex]\[ 3A = 6 \Rightarrow A = 2 \][/tex]
Substitute [tex]\( A = 2 \)[/tex] into equation (1):
[tex]\[ 2 + B = 1 \Rightarrow B = -1 \][/tex]
So, the partial fraction decomposition is:
[tex]\[ \frac{2}{x} + \frac{(-1)x - 1}{x^2 + 3} \][/tex]
[tex]\[ = \frac{2}{x} - \frac{x + 1}{x^2 + 3} \][/tex]
### Step 2: Integrate the Partial Fractions
Now, we integrate each term separately:
[tex]\[ \int \left( \frac{2}{x} - \frac{x + 1}{x^2 + 3} \right) dx \][/tex]
First, integrate [tex]\( \frac{2}{x} \)[/tex]:
[tex]\[ \int \frac{2}{x} dx = 2 \ln |x| \][/tex]
Next, integrate [tex]\( \frac{x + 1}{x^2 + 3} \)[/tex]. Break it into two integrals:
[tex]\[ \int \frac{x + 1}{x^2 + 3} dx = \int \frac{x}{x^2 + 3} dx + \int \frac{1}{x^2 + 3} dx \][/tex]
For [tex]\( \int \frac{x}{x^2 + 3} dx \)[/tex]:
Use the substitution [tex]\( u = x^2 + 3 \)[/tex], then [tex]\( du = 2x \, dx \)[/tex]:
[tex]\[ \int \frac{x}{x^2 + 3} dx = \frac{1}{2} \int \frac{2x}{x^2 + 3} dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| = \frac{1}{2} \ln|x^2 + 3| \][/tex]
For [tex]\( \int \frac{1}{x^2 + 3} dx \)[/tex]:
Recall the integral of [tex]\( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan \left( \frac{x}{a} \right) \)[/tex]:
[tex]\[ \int \frac{1}{x^2 + 3} dx = \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
Combining these integrals, we get:
[tex]\[ \int \frac{x + 1}{x^2 + 3} dx = \frac{1}{2} \ln|x^2 + 3| + \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
Thus, the integral becomes:
[tex]\[ \int \left( \frac{2}{x} - \frac{x + 1}{x^2 + 3} \right) dx \][/tex]
[tex]\[ = 2 \ln |x| - \left( \frac{1}{2} \ln |x^2 + 3| + \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \right) \][/tex]
[tex]\[ = 2 \ln |x| - \frac{1}{2} \ln |x^2 + 3| - \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
### Final Answer
Thus, the partial fraction decomposition of:
[tex]\[ \frac{x^2 - x + 6}{x^3 + 3x} \][/tex]
is:
[tex]\[ \frac{2}{x} - \frac{x + 1}{x^2 + 3} \][/tex]
And the integral is:
[tex]\[ \int \frac{x^2 - x + 6}{x^3 + 3x} dx = 2 \ln |x| - \frac{1}{2} \ln |x^2 + 3| - \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) + C \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
