Answer :
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]
