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The integral provided is:

\[
\int \frac{2x^2 + 5x + 3}{(x - 1)^2 (x^2 + 4)} \, dx
\]

To solve this integral, we can approach it using partial fraction decomposition and some algebraic manipulation. Let's break it down step by step:

### Step 1: Set up the Partial Fraction Decomposition

We express the integrand as a sum of simpler fractions. Because the denominator consists of two irreducible factors \((x - 1)^2\) and \((x^2 + 4)\), the partial fraction decomposition should have the following form:

\[
\frac{2x^2 + 5x + 3}{(x - 1)^2 (x^2 + 4)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{Cx + D}{x^2 + 4}
\]

Here:
- \( A \), \( B \), \( C \), and \( D \) are constants to be determined.

### Step 2: Multiply through by the common denominator

Multiply both sides of the equation by \((x - 1)^2 (x^2 + 4)\) to eliminate the denominator on the right-hand side:

\[
2x^2 + 5x + 3 = A(x - 1)(x^2 + 4) + B(x^2 + 4) + (Cx + D)(x - 1)^2
\]

### Step 3: Expand and collect like terms

Expand the right-hand side by distributing the terms, then collect powers of \(x\). This will give you a polynomial equation where you can match the coefficients of powers of \(x\) to solve for \(A\), \(B\), \(C\), and \(D\).

### Step 4: Solve for the constants

This will yield a system of equations that can be solved to determine the values of \(A\), \(B\), \(C\), and \(D\).

Once we have the constants, the integral can be broken down into simpler integrals, each of which can be solved using standard techniques such as:
- Basic integration for rational functions.
- Trigonometric substitution for integrals involving quadratic expressions.

I'll work through solving for the constants now.from sympy import symbols, Eq, solve, expand

# Define the variables and constants
x, A, B, C, D = symbols('x A B C D')

# Define the left-hand side and right-hand side of the equation after multiplying by common denominator
lhs = 2*x**2 + 5*x + 3
rhs = A*(x - 1)*(x**2 + 4) + B*(x**2 + 4) + (C*x + D)*(x - 1)**2

# Expand the right-hand side
rhs_expanded = expand(rhs)

# Create the equation lhs = rhs_expanded
equation = Eq(lhs, rhs_expanded)

# Solve the equation by matching coefficients of powers of x
solution = solve(equation, (A, B, C, D))

solutionThe partial fraction decomposition constants have been determined as:

\[
A = 1, \quad B = 2, \quad C = -1, \quad D = -1
\]

Thus, the decomposition of the original integrand is:

\[
\frac{2x^2 + 5x + 3}{(x - 1)^2(x^2 + 4)} = \frac{1}{x - 1} + \frac{2}{(x - 1)^2} - \frac{x + 1}{x^2 + 4}
\]

### Step 5: Solve the integrals separately

We now integrate each term:

1. \( \int \frac{1}{x - 1} \, dx = \ln|x - 1| \)
2. \( \int \frac{2}{(x - 1)^2} \, dx = -\frac{2}{x - 1} \)
3. \( \int \frac{-x}{x^2 + 4} \, dx = -\frac{1}{2} \ln(x^2 + 4) \) (using substitution)
4. \( \int \frac{-1}{x^2 + 4} \, dx = -\frac{1}{2} \arctan\left(\frac{x}{2}\right) \)

### Final Solution

Combining all these results, the final integral is:

\[
\int \frac{2x^2 + 5x + 3}{(x - 1)^2(x^2 + 4)} \, dx = \ln|x - 1| - \frac{2}{x - 1} - \frac{1}{2} \ln(x^2 + 4) - \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C
\]

where \(C\) is the constant of integration.

Let me know if you want further clarification or details!

### Relative Questions
1. How do you perform partial fraction decomposition for different types of rational functions?
2. What is the significance of trigonometric substitution in solving integrals?
3. How do you solve integrals involving quadratic expressions?
4. Why is it important to break down complex integrals into simpler parts?
5. What other methods can be used to solve integrals involving products of polynomials and quadratics?

### Tip
When dealing with rational functions, always check if the degree of the numerator is less than the denominator. If not, perform polynomial division first before attempting partial fraction decomposition.

Learn how to solve the integral of (2x^2 + 5x + 3) / ((x - 1)^2 (x^2 + 4)) step-by-step with detailed partial fraction decomposition and integration techniques. This problem involves advanced calculus concepts suitable for undergraduate students.

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