3.5 Q-27

Question Statement

Evaluate the integral:

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

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Background and Explanation

In this problem, we are tasked with solving an integral involving rational functions with quadratic denominators. To solve such integrals, we typically decompose the integrand into partial fractions. In this case, the denominator is factored into two quadratic expressions: $(x^2 + 4)$ and $(x^2 + 4x + 5)$. The method of partial fraction decomposition will allow us to break this complicated rational function into simpler fractions that are easier to integrate.

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Solution

To solve the integral, we first use partial fraction decomposition:

We start by assuming that the integrand can be written as the sum of two fractions:

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

Step 1: Multiply both sides by the denominator

Multiply both sides of the equation by $(x^2 + 4)(x^2 + 4x + 5)$ to eliminate the denominators:

$$4x + 1 = (Ax + B)(x^2 + 4x + 5) + (Cx + D)(x^2 + 4)$$

Step 2: Expand both sides

Now expand both sides:

$$4x + 1 = (Ax + B)(x^2 + 4x + 5) + (Cx + D)(x^2 + 4)$$

Expanding each term gives us:

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

Simplifying:

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

Step 3: Compare coefficients

Now, compare the coefficients of corresponding powers of $x$ on both sides of the equation:

• Coefficient of $x^3$: $A + C = 0$
• Coefficient of $x^2$: $4A + B + D = 0$
• Coefficient of $x$: $5A + 4B + 4C = 4$
• Constant term: $5B + 4D = 1$

Step 4: Solve the system of equations

We now solve the system of equations for $A$, $B$, $C$, and $D$:

1. From $A + C = 0$, we get $C = -A$.
2. From $5B + 4D = 1$, solve for $D$ in terms of $B$: $D = \frac{1 - 5B}{4}$
3. Substitute $C = -A$ and the expression for $D$ into the other two equations and solve:
   • $4A + B + D = 0$ becomes: $4A + B + \frac{1 - 5B}{4} = 0$
   • $5A + 4B + 4C = 4$ becomes: $5A + 4B + 4(-A) = 4$

Step 5: Final solution for coefficients

By solving the system, we find the following values for the coefficients:

• $A = 0$
• $B = 1$
• $C = 0$
• $D = -1$

Thus, the partial fraction decomposition simplifies to:

$$\frac{4x + 1}{(x^2 + 4)(x^2 + 4x + 5)} = \frac{1}{x^2 + 4} - \frac{1}{(x + 2)^2 + 1}$$

Step 6: Integrate

Now, we can integrate each term separately:

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

The first integral is a standard arctangent integral:

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

The second integral is also a standard arctangent integral:

$$\int \frac{1}{(x + 2)^2 + 1} , dx = \tan^{-1}(x + 2)$$

Thus, the solution to the original integral is:

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

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Key Formulas or Methods Used

• Partial Fraction Decomposition: Breaking a complicated rational function into simpler fractions.
• Arctangent Integral: The integrals of the forms $\int \frac{1}{x^2 + a^2} , dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$ and $\int \frac{1}{(x + b)^2 + 1} , dx = \tan^{-1}(x + b)$.

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Summary of Steps

1. Set up the partial fraction decomposition: $\frac{4x + 1}{(x^2 + 4)(x^2 + 4x + 5)} = \frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{x^2 + 4x + 5}$
2. Multiply both sides by the denominator and expand.
3. Compare the coefficients of powers of $x$ to form a system of equations.
4. Solve for $A$, $B$, $C$, and $D$.
5. Substitute the values of the coefficients into the decomposed expression.
6. Integrate each term using standard arctangent integrals.
7. Combine the results to get the final answer: $\frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) - \tan^{-1}(x + 2) + C$