Question

Evaluate the following integrals, which have irreducible quadratic factors. \(\int \frac{2}{(x-4)\left(x^{2}+2 x+6\right)} d x\)

Short answer

The integral evaluates to \(\frac{1}{15} \ln |x-4| + C_1 - \frac{1}{30} \ln |(x+1)^2 + 5| - \frac{2}{5\sqrt{5}} \arctan\left(\frac{x+1}{\sqrt{5}}\right) + C_2\).

Step-by-step solution

Setup Partial Fraction Decomposition

To evaluate the integral \( \int \frac{2}{(x-4)(x^{2}+2x+6)} \, dx \), we need to break it into partial fractions. Since \( x^2 + 2x + 6 \) is irreducible, the decomposition is: \[ \frac{2}{(x-4)(x^2+2x+6)} = \frac{A}{x-4} + \frac{Bx+C}{x^2+2x+6} \] where \(A, B, \) and \(C\) are constants to be determined.

Solve for Constants in the Partial Fraction Decomposition

Multiply through by the denominator \((x-4)(x^2+2x+6)\) to clear the fractions: \[ 2 = A(x^2 + 2x + 6) + (Bx + C)(x - 4) \] Expand and collect like terms:\[ 2 = Ax^2 + 2Ax + 6A + Bx^2 - 4Bx + Cx - 4C \] Combine terms:\[ 2 = (A + B)x^2 + (2A - 4B + C)x + (6A - 4C) \] To find \(A, B,\) and \(C\), equate coefficients with the left-hand side \(2\).

Determine Constants

Matching coefficients, we get three equations:1. \( A + B = 0 \)2. \( 2A - 4B + C = 0 \)3. \( 6A - 4C = 2 \)Solve this system of linear equations:- From equation (1): \( B = -A \) - Substitute into (2): \( 2A + 4A + C = 0 \Rightarrow C = -6A \)- Substitute \( B \) and \( C \) into (3): \( 6A - 4(-6A) = 2 \Rightarrow 30A = 2 \Rightarrow A = \frac{1}{15} \)Thus, \( B = -\frac{1}{15} \) and \( C = -\frac{2}{5} \).

Rewrite the Integral Using Partial Fractions

Now substitute \(A, B, \) and \(C\) back into the partial fractions: \[ \int \frac{2}{(x-4)(x^2+2x+6)} \, dx = \int \frac{1/15}{x-4} \, dx + \int \frac{-\frac{1}{15}x - \frac{2}{5}}{x^2+2x+6} \, dx \] Break the integral into two parts.

Solve Each Integral

The first integral is straightforward:\[ \int \frac{1/15}{x-4} \, dx = \frac{1}{15} \ln |x-4| + C_1 \]For the second integral, use substitution: complete the square for \(x^2 + 2x + 6\) to get \((x+1)^2 + 5\).Use \(u = x + 1\), \(du = dx\):\[ \int \frac{-\frac{1}{15}u - \frac{2}{5}}{u^2 + 5} \, du \]Split further as constant and linear terms and solve using standard integral results.

Final Integration and Combine Results

Solving the second integral gives:\[ -\frac{1}{15} \int \frac{u}{u^2+5} \, du - \frac{2}{5} \int \frac{1}{u^2+5} \, du \]The first part: \(-\frac{1}{15} \cdot \frac{1}{2} \ln |u^2+5| \), and for \(\int \frac{1}{u^2+5} du \) use arctangent:\[-\frac{2}{5} \cdot \frac{1}{\sqrt{5}} \arctan\left(\frac{u}{\sqrt{5}}\right) + C_2\]Substitute back \(u = x + 1\) and combine with the first integral result.

Key concepts

Integration Techniques

Integrating functions can sometimes be complex, especially when dealing with rational expressions like \( \int \frac{2}{(x-4)(x^2+2x+6)} \, dx \). To simplify this process, we use various integration techniques. One powerful technique is Partial Fraction Decomposition, which is used to break complex fractions into simpler components. This allows each part to be integrated separately. Once decomposed, you end up with simpler integrals, like linear over linear or linear over quadratic, which can often be solved using basic integral formulas or substitutions.

In our example, we first rearrange the integrand using the decomposition into partial fractions, which transforms our integral into a sum of simpler fractions. Each fraction can then be tackled on its own using basic integration formulas or substitutions designed for those specific types of functions. By approaching the integration in manageable chunks, we make the overall solution more straightforward and efficient.
Another common technique used in conjunction with this is substitution, especially when the integral involves irreducible quadratic polynomials. Substitutions can simplify integrals by changing variables to forms that are easier to integrate.

Irreducible Quadratic

When working with partial fractions, you often come across quadratic expressions. If a quadratic factor does not factor further into linear terms over the real numbers, it is termed as irreducible. One example is the quadratic \( x^2 + 2x + 6 \), which appears in our exercise. To integrate functions involving irreducible quadratics, you typically set up your partial fractions' decomposition to include a term in the form \( \frac{Bx+C}{x^2+2x+6} \).

In the denominator of this fraction, \( x^2 + 2x + 6 \), attempts to factor it into real number components would not succeed as it doesn't have real roots. Recognizing this means that one must use alternative techniques like completing the square or substitution to simplify the integration. Completing the square for the quadratic gives us a more convenient expression when making substitutions. In this case, completing the square transforms \( x^2 + 2x + 6 \) to \( (x+1)^2 + 5 \). This puts the denominator in a recognizable form useful for integration.
Remember, knowing when a quadratic is irreducible and using the appropriate techniques to work with it is crucial for solving these kinds of integrals.

Linear Equations

Solving systems of linear equations is a key step in partial fraction decomposition. After setting up the decomposition, we find constants by solving equations that arise from matching coefficients of polynomials. For instance, when you have \[ (A + B)x^2 + (2A - 4B + C)x + (6A - 4C) = 2 \], you compare coefficients of like terms on both sides of the equation.

In our exercise, we saw three resulting equations from comparing coefficients:

• \( A + B = 0 \)
• \( 2A - 4B + C = 0 \)
• \( 6A - 4C = 2 \)

Solving this system involves simple algebraic techniques such as substitution or elimination. By solving these, you end up with specific values for \( A \), \( B \), and \( C \).
This process heavily demonstrates the interconnectedness of different mathematical skills—from setting up correct equations, switching expressions during substitution, to solving basic linear equations all aimed at simplifying more complex integrals.

Completing the Square

Completing the square is not only crucial in vertex form equations but also in calculus, particularly when dealing with partial fraction decomposition involving irreducible quadratics. It helps convert quadratic expressions into a form that is easier to integrate.

Given the quadratic \( x^2 + 2x + 6 \), completing the square involves rewriting it as \( (x+1)^2 + 5 \). Here’s how it works: start by taking half of the coefficient of \( x \) (which is 2), squaring it to get 1, and then add and subtract this value within the same expression. Though completing the square may seem tedious, it is quite helpful. In integration, especially for arcsine and arctan forms, this transformation is critical.
Once the quadratic is in the squared form plus a constant, it sets the stage for applying integration techniques, such as trigonometrical substitutions or standard integrals involving \( u^2 + a^2 \).
This technique simplifies complex problems into relatable and solvable forms, illustrating the profound utility of algebraic skills in calculus.