Problem 65 - Calculus Single Variable

Question

Calculate the given integral. $$ \int \frac{6 x}{x^{2}-14 x+58} d x $$

Step-by-Step Solution

Answer

\( 3 \log|x^2 - 14x + 58| + 14 \arctan\left( \frac{x-7}{3} \right) + C \)

Step 1: Identify the integral form

The given integral is \[ \int \frac{6x}{x^2 - 14x + 58} \, dx \].We notice that the denominator is a quadratic expression, indicating a potential substitution method or completing the square might be helpful.

Step 2: Complete the square for the denominator

The quadratic expression in the denominator is \( x^2 - 14x + 58 \).To complete the square, we rearrange it as follows:\[ x^2 - 14x + 58 = (x-7)^2 + 9 \].Thus, the integral can be rewritten as\[ \int \frac{6x}{(x-7)^2 + 9} \, dx \].

Step 3: Use substitution to simplify the integral

Let \( u = x - 7 \), then \( du = dx \) and \( x = u + 7 \).Substituting these into the integral, we have\[ \int \frac{6(u+7)}{u^2 + 9} \, du \].This expression can be split into two separate integrals:\[ \int \frac{6u}{u^2 + 9} \, du + \int \frac{42}{u^2 + 9} \, du \].

Step 4: Solve the first integral

For the first integral, \( \int \frac{6u}{u^2 + 9} \, du \), use substitution:Let \( v = u^2 + 9 \), then \( dv = 2u \, du \).Thus, \( 3 \int \frac{2u}{u^2 + 9} \, du = 3 \int \frac{1}{v} \, dv \), which evaluates to\[ 3 \log|v| = 3 \log|u^2 + 9| \].

Step 5: Solve the second integral

The second integral, \( \int \frac{42}{u^2 + 9} \, du \), is a standard form that translates to\( \frac{42}{3} \int \frac{1}{(u/3)^2 + 1} \, du \),which results in\[ 14 \arctan\left( \frac{u}{3} \right) \].

Step 6: Combine the results and back-substitute

Combine the results of the two integrals:\[ 3 \log|u^2 + 9| + 14 \arctan\left( \frac{u}{3} \right) + C \].Substitute \( u = x - 7 \) back into the expression:\[ 3 \log|x^2 - 14x + 58| + 14 \arctan\left( \frac{x-7}{3} \right) + C \].

Key Concepts

Substitution Method

The substitution method is a powerful tool in evaluating integrals. It's all about making a substitution that simplifies the integral. We start by choosing a new variable, say \( u \), which seems to align with part of the integral's expression. By changing variables, the integral often becomes easier to solve.
In our exercise, we used substitution twice. First, we set \( u = x - 7 \) so our quadratic expression becomes square-completed, simplifying the denominator. After this, another substitution \( v = u^2 + 9 \) was employed. This transformed another part of the integral into a simple logarithmic form: \( \ \int \frac{1}{v} \, dv \), which we know is \( \ \log|v| + C \).
Key tips for using substitution effectively include:

• Always adjust the differential \( dx \) to match your new variable, \( du \).
• Don't forget to transform the limits of integration if you are working with definite integrals.
• Substitute back the original variable to arrive at the final solution.

Completing the Square

Completing the square is essential for simplifying quadratic expressions, making them easier to integrate. It involves adding and subtracting terms to transform the quadratic form \( ax^2 + bx + c \) into \( (x-h)^2 + k \). This form reveals a perfect square trinomial along with a constant, making it manageable for integration.
For example, in our integral, \( x^2 - 14x + 58 \) was transformed into \( (x-7)^2 + 9 \). This simplification enabled easier handling via substitution.
When completing the square:

• Rewrite the quadratic expression by focusing on the terms that form the square.
• Divide the coefficient of \( x \) by 2, square it, and add and subtract it within the expression.
• Express the rearranged terms as a perfect square \((x-h)^2\) plus or minus a constant.

Arctangent Integration

The arctangent integration technique is used for integrals that have a form reminiscent of \( \ \int \frac{1}{x^2 + a^2} \, dx \). This integral resolves into an arctangent function, specifically, \( \ rac{1}{a} \arctan\left( \frac{x}{a} \right) + C \).
We applied this in solving \( \ \int \frac{42}{u^2 + 9} \, du \) in our exercise. Recognizing the form allowed us to use \( a = 3 \), giving us the solution \( 14 \arctan\left( \frac{u}{3} \right) \).
Key points when using arctangent integration:

• Make sure your integrand matches the \( \frac{1}{x^2 + a^2} \) form. Adjust coefficients as needed.
• Remember the factor \( \frac{1}{a} \) before the arctangent function.
• Account for any transformations such as substitutions affecting the variable \( x \).

Logarithmic Integration

Logarithmic integration arises mainly when dealing with integrals of the form \( \ \int \frac{f'(x)}{f(x)} \, dx \), which simplifies to \( \ \log|f(x)| + C \). Sometimes recognized by a straightforward \( \ \int \frac{1}{x} \, dx \), which gives \( \ \log|x| + C \).
In our example, the substitution \( v = u^2 + 9 \) simplified \( \ \int \frac{6u}{u^2 + 9} \, du \) to \( \ 3 \int \frac{1}{v} \, dv \), resulting in \( \ 3 \log|v| \).
Useful hints for employing logarithmic integration:

• Ensure the derivative \( f'(x) \) is present in the numerator or can be factored out.
• If the expression isn't quite right, try substitution or algebraic manipulation.
• After integrating, return to your original variables if substitutions were involved.