IDNLearn.com offers a comprehensive solution for all your question and answer needs. Our community provides accurate and timely answers to help you understand and solve any issue.
Sagot :
Certainly! Let's solve the given integral step by step.
We start by performing a partial fraction decomposition of the integrand.
### Step 1: Partial Fraction Decomposition
Given the integral:
[tex]\[ \int \frac{3x+5}{(x+1)(x-1)^2} \, dx \][/tex]
We want to decompose [tex]\(\frac{3x+5}{(x+1)(x-1)^2}\)[/tex] into simpler fractions. The appropriate form for the partial fraction decomposition is:
[tex]\[ \frac{3x+5}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2} \][/tex]
### Step 2: Finding Coefficients [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]
To find the constants [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex], we write:
[tex]\[ 3x + 5 = A(x-1)^2 + B(x+1) + C(x+1)(x-1) \][/tex]
Expanding and simplifying:
[tex]\[ 3x + 5 = A(x^2 - 2x + 1) + B(x + 1) + C(x^2 - 1) \][/tex]
[tex]\[ 3x + 5 = Ax^2 - 2Ax + A + Bx + B + Cx^2 - C \][/tex]
[tex]\[ 3x + 5 = (A + C)x^2 + (-2A + B)x + (A + B - C) \][/tex]
By equating coefficients of [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant term, we get three equations:
1. For [tex]\(x^2\)[/tex]: [tex]\(A + C = 0\)[/tex]
2. For [tex]\(x\)[/tex]: [tex]\(-2A + B = 3\)[/tex]
3. For constants: [tex]\(A + B - C = 5\)[/tex]
From equation (1):
[tex]\[ C = -A \][/tex]
Substitute [tex]\(C = -A\)[/tex] into equations (2) and (3):
[tex]\[ -2A + B = 3 \quad \text{(2)} \][/tex]
[tex]\[ A + B - (-A) = 5 \quad \text{(3)} \][/tex]
Simplifying equation (3):
[tex]\[ A + B + A = 5 \][/tex]
[tex]\[ 2A + B = 5 \quad \text{(3)} \][/tex]
Now we have:
1. [tex]\(-2A + B = 3\)[/tex]
2. [tex]\(2A + B = 5\)[/tex]
Subtract equation (1) from equation (2):
[tex]\[ [(2A + B) - (-2A + B)] = 5 - 3 \][/tex]
[tex]\[ 4A = 2 \][/tex]
[tex]\[ A = \frac{1}{2} \][/tex]
[tex]\[ C = -A = -\frac{1}{2} \][/tex]
Now substitute [tex]\(A = \frac{1}{2}\)[/tex] into either equation to find [tex]\(B\)[/tex]:
[tex]\[ 2(\frac{1}{2}) + B = 5 \][/tex]
[tex]\[ 1 + B = 5 \][/tex]
[tex]\[ B = 4 \][/tex]
So, we have:
[tex]\[ A = \frac{1}{2}, \quad B = 4, \quad C = -\frac{1}{2} \][/tex]
Thus the partial fraction decomposition is:
[tex]\[ \frac{3x+5}{(x+1)(x-1)^2} = \frac{\frac{1}{2}}{x+1} + \frac{4}{x-1} - \frac{\frac{1}{2}}{(x-1)^2} \][/tex]
### Step 3: Integrating Each Term
Now, we integrate each term separately:
[tex]\[ \int \frac{3x+5}{(x+1)(x-1)^2} \, dx = \int \left( \frac{\frac{1}{2}}{x+1} + \frac{4}{x-1} - \frac{\frac{1}{2}}{(x-1)^2} \right) \, dx \][/tex]
[tex]\[ = \frac{1}{2} \int \frac{1}{x+1} \, dx + 4 \int \frac{1}{x-1} \, dx - \frac{1}{2} \int \frac{1}{(x-1)^2} \, dx \][/tex]
Integrating each term:
[tex]\[ \int \frac{1}{x+1} \, dx = \ln|x+1| + C_1 \][/tex]
[tex]\[ \int \frac{1}{x-1} \, dx = \ln|x-1| + C_2 \][/tex]
[tex]\[ \int \frac{1}{(x-1)^2} \, dx = -\frac{1}{x-1} + C_3 \][/tex]
Combining these, we get:
[tex]\[ \frac{1}{2} \ln|x+1| + 4 \ln|x-1| - \frac{1}{2} \left( -\frac{1}{x-1} \right) + C \][/tex]
Simplifying this, we have:
[tex]\[ \frac{1}{2} \ln|x+1| + 4 \ln|x-1| + \frac{1}{2(x-1)} + C \][/tex]
Using logarithm properties, we can combine the logarithms:
[tex]\[ \boxed{-\frac{1}{2} \ln|x-1| + \frac{1}{2} \ln|x+1| - \frac{4}{x-1} + C} \][/tex]
This is the final answer:
[tex]\[ \boxed{-\frac{1}{2} \ln|x-1| + \frac{1}{2} \ln|x+1| - \frac{4}{x-1} + C} \][/tex]
We start by performing a partial fraction decomposition of the integrand.
### Step 1: Partial Fraction Decomposition
Given the integral:
[tex]\[ \int \frac{3x+5}{(x+1)(x-1)^2} \, dx \][/tex]
We want to decompose [tex]\(\frac{3x+5}{(x+1)(x-1)^2}\)[/tex] into simpler fractions. The appropriate form for the partial fraction decomposition is:
[tex]\[ \frac{3x+5}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2} \][/tex]
### Step 2: Finding Coefficients [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]
To find the constants [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex], we write:
[tex]\[ 3x + 5 = A(x-1)^2 + B(x+1) + C(x+1)(x-1) \][/tex]
Expanding and simplifying:
[tex]\[ 3x + 5 = A(x^2 - 2x + 1) + B(x + 1) + C(x^2 - 1) \][/tex]
[tex]\[ 3x + 5 = Ax^2 - 2Ax + A + Bx + B + Cx^2 - C \][/tex]
[tex]\[ 3x + 5 = (A + C)x^2 + (-2A + B)x + (A + B - C) \][/tex]
By equating coefficients of [tex]\(x^2\)[/tex], [tex]\(x\)[/tex], and the constant term, we get three equations:
1. For [tex]\(x^2\)[/tex]: [tex]\(A + C = 0\)[/tex]
2. For [tex]\(x\)[/tex]: [tex]\(-2A + B = 3\)[/tex]
3. For constants: [tex]\(A + B - C = 5\)[/tex]
From equation (1):
[tex]\[ C = -A \][/tex]
Substitute [tex]\(C = -A\)[/tex] into equations (2) and (3):
[tex]\[ -2A + B = 3 \quad \text{(2)} \][/tex]
[tex]\[ A + B - (-A) = 5 \quad \text{(3)} \][/tex]
Simplifying equation (3):
[tex]\[ A + B + A = 5 \][/tex]
[tex]\[ 2A + B = 5 \quad \text{(3)} \][/tex]
Now we have:
1. [tex]\(-2A + B = 3\)[/tex]
2. [tex]\(2A + B = 5\)[/tex]
Subtract equation (1) from equation (2):
[tex]\[ [(2A + B) - (-2A + B)] = 5 - 3 \][/tex]
[tex]\[ 4A = 2 \][/tex]
[tex]\[ A = \frac{1}{2} \][/tex]
[tex]\[ C = -A = -\frac{1}{2} \][/tex]
Now substitute [tex]\(A = \frac{1}{2}\)[/tex] into either equation to find [tex]\(B\)[/tex]:
[tex]\[ 2(\frac{1}{2}) + B = 5 \][/tex]
[tex]\[ 1 + B = 5 \][/tex]
[tex]\[ B = 4 \][/tex]
So, we have:
[tex]\[ A = \frac{1}{2}, \quad B = 4, \quad C = -\frac{1}{2} \][/tex]
Thus the partial fraction decomposition is:
[tex]\[ \frac{3x+5}{(x+1)(x-1)^2} = \frac{\frac{1}{2}}{x+1} + \frac{4}{x-1} - \frac{\frac{1}{2}}{(x-1)^2} \][/tex]
### Step 3: Integrating Each Term
Now, we integrate each term separately:
[tex]\[ \int \frac{3x+5}{(x+1)(x-1)^2} \, dx = \int \left( \frac{\frac{1}{2}}{x+1} + \frac{4}{x-1} - \frac{\frac{1}{2}}{(x-1)^2} \right) \, dx \][/tex]
[tex]\[ = \frac{1}{2} \int \frac{1}{x+1} \, dx + 4 \int \frac{1}{x-1} \, dx - \frac{1}{2} \int \frac{1}{(x-1)^2} \, dx \][/tex]
Integrating each term:
[tex]\[ \int \frac{1}{x+1} \, dx = \ln|x+1| + C_1 \][/tex]
[tex]\[ \int \frac{1}{x-1} \, dx = \ln|x-1| + C_2 \][/tex]
[tex]\[ \int \frac{1}{(x-1)^2} \, dx = -\frac{1}{x-1} + C_3 \][/tex]
Combining these, we get:
[tex]\[ \frac{1}{2} \ln|x+1| + 4 \ln|x-1| - \frac{1}{2} \left( -\frac{1}{x-1} \right) + C \][/tex]
Simplifying this, we have:
[tex]\[ \frac{1}{2} \ln|x+1| + 4 \ln|x-1| + \frac{1}{2(x-1)} + C \][/tex]
Using logarithm properties, we can combine the logarithms:
[tex]\[ \boxed{-\frac{1}{2} \ln|x-1| + \frac{1}{2} \ln|x+1| - \frac{4}{x-1} + C} \][/tex]
This is the final answer:
[tex]\[ \boxed{-\frac{1}{2} \ln|x-1| + \frac{1}{2} \ln|x+1| - \frac{4}{x-1} + C} \][/tex]
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.
