Problem 58

Question

Evaluate the following integrals. $$\int \frac{d x}{x^{2}-6 x+34}$$

Step-by-Step Solution

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Answer

Answer: The antiderivative of the given integral is $\arctan\left(\frac{x-3}{5}\right) + C$.

1Step 1: Complete the square on the denominator

In the denominator we have $x^2 - 6x + 34$. Completing the square on the quadratic expression, we add and subtract $(-6/2)^2 = 9$ to maintain equality: $(x^2 - 6x + 9) + 25 = (x - 3)^2 + 5^2$ Now, the integral becomes $$\int \frac{dx}{(x-3)^2 + 5^2}$$

2Step 2: Identify the standard form of the integral and use a suitable substitution

Now, we can see that this integral is in the standard form of $$\int \frac{dx}{a^2 + x^2}$$ where $a = 5$. This form of integral has the antiderivative $\frac{1}{a}\arctan(\frac{x}{a}) + C$. To make the substitution, let $x - 3 = 5\tan(\theta)$, so $dx = 5\sec^2(\theta)d\theta$ and we have $$\int \frac{5\sec^2(\theta)d\theta}{5^2+(\tan(\theta))^2\cdot5^2}$$

3Step 3: Simplify and integrate

The integral now simplifies to $$\int \frac{\sec^2(\theta)}{1+\tan^2(\theta)}d\theta$$ Using the trigonometric identity $1+\tan^2(\theta) = \sec^2(\theta)$, the integral simplifies further: $$\int d\theta=\theta + C$$

4Step 4: Substitute back the original variables

Now, let's return back to our original substitution. Since $x - 3 = 5\tan(\theta)$, we have $\theta = \arctan(\frac{x-3}{5})$. Thus, the antiderivative of the given integral is $$\arctan \left(\frac{x-3}{5}\right) + C$$

Key Concepts

Completing the SquareStandard Integral FormsTrigonometric SubstitutionAntiderivative

Completing the Square

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This simplifies the quadratic into a form that is easier to integrate. In the original exercise, we start with the quadratic polynomial in the form of:

$x^2 - 6x + 34$

To complete the square, identify the coefficient of the linear term, which is -6, divide it by 2, and then square it:

$(-6/2)^2 = 9$

Add and subtract 9 inside the expression to maintain equality:

$(x^2 - 6x + 9) + 25$

This results in the perfect square

$(x - 3)^2 + 25$

This transformation simplifies the integration process by converting the denominator into a recognizable standard form.

Standard Integral Forms

Standard integral forms are key tools in solving integrals that conform to a certain pattern or "template". Recognizing these forms can greatly simplify the integration process. After completing the square, the integral becomes:

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

This matches the standard form of $ \int \frac{dx}{a^2 + x^2} $, where $ a=5 $. The solution to this standard form is a widely known antiderivative:

$ \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C $

Recognizing this allows us to avoid complicated integration and immediately move towards solving the integral using this well-known template.

Trigonometric Substitution

Trigonometric substitution is a technique used in calculus to simplify the integration of functions involving square roots and certain rational expressions. In our example, following the transformation using the standard form, we make the substitution

$ x - 3 = 5\tan(\theta) $

This implies:

$ dx = 5\sec^2(\theta)d\theta $

Substituting these into the integral, we transform it further:

$ \int \frac{5\sec^2(\theta)d\theta}{25+25\tan^2(\theta)} $

Using the trigonometric identity $ 1 + \tan^2(\theta) = \sec^2(\theta) $, the integral simplifies significantly:

$ \int d\theta $

Thus, trigonometric substitution changes a complicated expression into an easily integrable form.

Antiderivative

The antiderivative, or indefinite integral, is the function that reverses the process of differentiation. It represents a family of functions whose derivatives are given by a specific function. In solving the given integral, after simplifying the expression, we arrive at: