3.5 Q-24

Question Statement

Evaluate the integral:

$$\int \frac{2 x^{2}+5 x+3}{(x-1)^{2}\left(x^{2}+4\right)} dx$$

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Background and Explanation

To solve this integral, we’ll use partial fraction decomposition. The goal is to break the rational function into simpler fractions that are easier to integrate. We will split the fraction into terms involving $(x-1)$ and $(x^2+4)$, as the denominator is a product of these factors. We need to determine the constants for each term in the decomposition.

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Solution

Step 1: Set up the partial fraction decomposition

We start by assuming the form of the decomposition:

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

We will solve for the constants $A$, $B$, $C$, and $D$.

Step 2: Multiply both sides by the denominator

Multiply both sides of the equation by $(x-1)^2(x^2+4)$ to clear the fractions:

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

Step 3: Substitute $x = 1$ to solve for $B$

To simplify the process of solving for the constants, substitute $x = 1$ into the equation:

$$2(1)^2 + 5(1) + 3 = B((1)^2 + 4)$$

This simplifies to:

$$2 + 5 + 3 = B(1 + 4) \quad \Rightarrow \quad B = \frac{10}{5} = 2$$

Step 4: Compare coefficients

Now, expand both sides and compare the coefficients of like powers of $x$. First, expand both sides of the equation:

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

From here, we can compare the coefficients of $x^3$, $x^2$, $x$, and the constant term to form a system of equations. The results give:

• $A + C = 0$
• $-A + B - 2C + D = 2$
• $-4A + 4B + D = 3$
• $-A + 2C + D = 2$

Step 5: Solve the system of equations

Using the previously found value $B = 2$, and solving the system:

• $A + C = 0 \quad \Rightarrow A = -C$
• Substituting $A = -C$ into the equations and solving gives $C = -1$, $A = 1$, and $D = -1$.

Step 6: Write the integral in terms of simpler integrals

Now that we have $A = 1$, $B = 2$, $C = -1$, and $D = -1$, we can rewrite the integral:

$$\int \frac{2 x^2 + 5 x + 3}{(x-1)^2 (x^2 + 4)} dx = \int \left( \frac{1}{x-1} + \frac{2}{(x-1)^2} - \frac{x+1}{x^2 + 4} \right) dx$$

Step 7: Integrate each term

Now, we integrate each term individually:

• $\int \frac{1}{x-1} dx = \ln |x-1|$

• $\int \frac{2}{(x-1)^2} dx = -\frac{2}{x-1}$

• $\int \frac{x+1}{x^2 + 4} dx$ can be split into two parts: $\int \frac{x}{x^2 + 4} dx + \int \frac{1}{x^2 + 4} dx$

  The first integral is straightforward: $\int \frac{x}{x^2 + 4} dx = \frac{1}{2} \ln(x^2 + 4)$

  The second integral is: $\int \frac{1}{x^2 + 4} dx = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right)$

Step 8: Combine the results

Putting everything together, the final solution is:

$$\ln |x-1| - \frac{2}{x-1} - \frac{1}{2} \ln(x^2 + 4) - \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C$$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: Breaking a rational function into simpler fractions.
• Logarithmic Integration: Integrals of the form $\frac{1}{x-a}$.
• Arctangent Formula: $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$

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Summary of Steps

1. Set up the partial fraction decomposition: $\frac{2x^2 + 5x + 3}{(x-1)^2(x^2 + 4)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx + D}{x^2 + 4}$

2. Multiply both sides by the denominator and substitute $x = 1$ to find $B = 2$.

3. Compare coefficients of like powers of $x$ to find $A = 1$, $C = -1$, and $D = -1$.

4. Rewrite the integral in simpler terms: $\int \left( \frac{1}{x-1} + \frac{2}{(x-1)^2} - \frac{x+1}{x^2 + 4} \right) dx$

5. Integrate each term:

   • $\int \frac{1}{x-1} dx = \ln |x-1|$
   • $\int \frac{2}{(x-1)^2} dx = -\frac{2}{x-1}$
   • $\int \frac{x+1}{x^2 + 4} dx = \frac{1}{2} \ln(x^2 + 4) + \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right)$

6. Combine the results and write the final answer: $\ln |x-1| - \frac{2}{x-1} - \frac{1}{2} \ln(x^2 + 4) - \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C$