Question

Evaluate the following integrals. $$\int \frac{d x}{2 x^{2}-12 x+36}$$

Short answer

Question: Evaluate the following integral: $$\int \frac{dx}{2x^2 - 12x + 36}$$ Answer: $$\int \frac{dx}{2x^2 - 12x + 36} = \frac{1}{18}\ln|x-3| + C$$

Step-by-step solution

Factor the denominator

Factor the quadratic in the denominator to see if it can be simplified: $$2x^2 - 12x + 36 = 2(x^2 - 6x + 18)$$ Since the quadratic inside the parenthesis cannot be factored further, we will use the expression as is.

Perform partial fraction decomposition

Now that we have factored the denominator, we will use partial fraction decomposition to rewrite the integrand. Let $$\frac{1}{2x^2 -12x + 36} = \frac{A}{x - h}$$ where h is some constant to be determined. Multiplying both sides by \((x - h)(2x^2 -12x + 36)\) gives $$1 = A(2x^2 -12x + 36)$$ Now we will complete the square to find the appropriate value of \(h\): $$x^2 - 6x + 18 = (x - 3)^2 + 9$$ So our integrand becomes: $$\frac{1}{2((x-3)^2 + 9)} = \frac{A}{x - 3}$$ Multiplying both sides by \((x - 3)(2((x-3)^2 + 9))\) gives $$1 = A(2((x-3)^2 + 9))$$ Comparing the coefficients, we get \(A = \frac{1}{18}\). Therefore our integrand can be written as: $$\int \frac{dx}{2x^2 - 12x + 36} = \int \frac{1/18}{x-3} dx$$

Evaluate the integral

Now we can easily evaluate the integral: $$\int \frac{\frac{1}{18}}{x-3} dx = \frac{1}{18}\int \frac{1}{x-3} dx$$ Using the integration rule \(\int \frac{1}{x} dx = \ln|x| + C\): $$\frac{1}{18}\int \frac{1}{x-3} dx = \frac{1}{18}\ln|x-3| + C$$ So, the final answer is: $$\int \frac{dx}{2x^2 - 12x + 36} = \frac{1}{18}\ln|x-3| + C$$

Key concepts

Partial Fractions

Partial fractions is a technique used to break down complex rational functions into simpler fractions that are easier to work with. It is particularly useful in integration when dealing with complicated denominators. Essentially, you decompose a fraction into a sum of simpler fractions, which can then be integrated individually. To use partial fraction decomposition effectively, you generally start by factoring the denominator, as shown in the original solution where the expression \(2x^2 - 12x + 36\) was factored. If the factorized form of the quadratic doesn't simplifiy further or isn't linear factors immediately, you can often complete the square or make another suitable move to find affected constants or rearrange your integrand.Here's how to perform partial fractions:

• Identify the form of the partial fraction based on the factors of the denominator.
• Set up the decomposed fraction with unknown coefficients.
• Multiply through by the common denominator to get rid of the fractions.
• Expand and equate the original numerator to the new numerator.
• Solve for the unknown coefficients by substituting convenient values for the variable or by solving the resulting system of equations.

Using partial fractions method in integrations simplifies computation, often changing complex expressions into linear terms that are directly integrable.

Integration by Substitution

Integration by substitution is a key technique in calculus used to simplify integrals and make them easier to solve by changing the variable. Essentially, it often involves a change of variables to transform a difficult integral into a much simpler one, thereby making it possible to apply known formulas or basic integration techniques.In the original problem, substitution isn't directly used, but an equivalent transformation occurs through factoring the quadratic expression and subsequent manipulation. Nonetheless, understanding substitution is vital in integration:

• Identify a part of the integral that can be substituted by a single variable (commonly \(u\)).
• Differentiate \(u\) to express \(dx\) in terms of \(du\).
• Replace the identified part of the integral and \(dx\) with \(u\) and \(du\).
• Perform the integration with respect to \(u\) using simpler formulas.
• Substitute \(u\) back into the original variable to get the final answer.

This technique often leads to elegant solutions by effectively managing the integration through simplification and transformation, and is especially useful when dealing with rational functions, trigonometric integrals, and exponential functions.

Quadratic Expressions

Quadratic expressions frequently appear in integrals, and knowing how to manipulate them is crucial for solving integrals effectively. These expressions take the general form \(ax^2 + bx + c\) and often require factorizing or completing the square to simplify.In the given solution, the expression \(2x^2 - 12x + 36\) was factored and then reformulated by completing the square. Completing the square is a method used to turn a quadratic expression into a perfect square trinomial plus a constant,Making our form easier to work with, such as for: \[x^2 - 6x + 18 = (x - 3)^2 + 9.\]This manipulation is particularly handy when dealing with quadratic denominators. Here's why understanding quadratic expressions is important:

• It helps in simplifying integrals by transforming them into forms that are more manageable, like those involving simpler substitutions.
• Extracting these forms is necessary for techniques like partial fraction decomposition.
• It aids in recognizing integrals that fit known results or standard forms, facilitating direct integration.

Being familiar with these expressions ensures efficiency in breaking down and evaluating complex integrals, leading to more straightforward solutions.