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To solve the integral [tex]\(\int \frac{6x + 2}{9x^2 + 6x + 26} \, dx\)[/tex], we can use a method involving substitution and simplification. Here's a step-by-step explanation:
### Step 1: Recognize the Form of the Integral
Observe that the integrand is a rational function where the numerator is a linear polynomial and the denominator is a quadratic polynomial. This suggests that we might be able to simplify the integrand by using a technique that involves recognizing the derivative of the denominator within the numerator.
### Step 2: Simplify the Integrand
The denominator of the integrand is [tex]\(9x^2 + 6x + 26\)[/tex]. Let's differentiate this with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{d}{dx}(9x^2 + 6x + 26) = 18x + 6 \][/tex]
Notice that the numerator [tex]\(6x + 2\)[/tex] can be related to [tex]\(18x + 6\)[/tex]. Specifically, [tex]\(6x + 2\)[/tex] is related but not directly a multiple of [tex]\(18x + 6\)[/tex].
### Step 3: Perform Substitution if Necessary
To help express [tex]\(6x + 2\)[/tex] more conveniently, let’s use:
[tex]\[ u = 9x^2 + 6x + 26 \\ du = (18x + 6) dx \][/tex]
But, we only have [tex]\(6x + 2\)[/tex] in the integrand, not [tex]\(18x + 6\)[/tex]. Therefore, it helps to think in terms of modifying the numerator slightly. By dividing and adjusting coefficients, we can map it back:
[tex]\[ (6x+2)dx = \frac{1}{3}(18x + 6)dx - \frac{2}{3} dx \][/tex]
This formulation is expressing [tex]\(6x + 2\)[/tex] as [tex]\(1/3(18x + 6)\)[/tex].
### Step 4: Rewrite the Integral
The integral can then be written as:
[tex]\[ \int \frac{6x + 2}{9x^2 + 6x + 26} \, dx = \int \frac{1/3 (18x + 6)}{9x^2 + 6x + 26} \, dx \][/tex]
Thus, we can now separate:
[tex]\[ = \frac{1}{3} \int \frac{18x + 6}{9x^2 + 6x + 26} \, dx \][/tex]
Rewriting this, we get a clearer form where each part is related to a derivative of the denominator:
[tex]\[ = \frac{1}{3} \int \frac{d(9x^2 + 6x + 26)}{9x^2 + 6x + 26} \][/tex]
### Step 5: Integrate Using Logarithm Rule
The integral of a function of the form [tex]\(\frac{du}{u}\)[/tex] is [tex]\(\ln|u|\)[/tex]. Therefore, we have:
[tex]\[ \int \frac{d(9x^2 + 6x + 26)}{9x^2 + 6x + 26} = \ln|9x^2 + 6x + 26| + C \][/tex]
Thus, our integral becomes:
[tex]\[ = \frac{1}{3} \ln|9x^2 + 6x + 26| + C \][/tex]
Since [tex]\(9x^2 + 6x + 26\)[/tex] is always positive for all [tex]\(x\)[/tex] (as it does not have real roots), the absolute value can be dropped:
[tex]\[ = \frac{1}{3} \ln(9x^2 + 6x + 26) + C \][/tex]
### Conclusion
Therefore, the solution to the integral [tex]\(\int \frac{6x + 2}{9x^2 + 6x + 26} \, dx\)[/tex] is:
[tex]\[ \boxed{\frac{1}{3} \ln(9x^2 + 6x + 26) + C} \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
### Step 1: Recognize the Form of the Integral
Observe that the integrand is a rational function where the numerator is a linear polynomial and the denominator is a quadratic polynomial. This suggests that we might be able to simplify the integrand by using a technique that involves recognizing the derivative of the denominator within the numerator.
### Step 2: Simplify the Integrand
The denominator of the integrand is [tex]\(9x^2 + 6x + 26\)[/tex]. Let's differentiate this with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{d}{dx}(9x^2 + 6x + 26) = 18x + 6 \][/tex]
Notice that the numerator [tex]\(6x + 2\)[/tex] can be related to [tex]\(18x + 6\)[/tex]. Specifically, [tex]\(6x + 2\)[/tex] is related but not directly a multiple of [tex]\(18x + 6\)[/tex].
### Step 3: Perform Substitution if Necessary
To help express [tex]\(6x + 2\)[/tex] more conveniently, let’s use:
[tex]\[ u = 9x^2 + 6x + 26 \\ du = (18x + 6) dx \][/tex]
But, we only have [tex]\(6x + 2\)[/tex] in the integrand, not [tex]\(18x + 6\)[/tex]. Therefore, it helps to think in terms of modifying the numerator slightly. By dividing and adjusting coefficients, we can map it back:
[tex]\[ (6x+2)dx = \frac{1}{3}(18x + 6)dx - \frac{2}{3} dx \][/tex]
This formulation is expressing [tex]\(6x + 2\)[/tex] as [tex]\(1/3(18x + 6)\)[/tex].
### Step 4: Rewrite the Integral
The integral can then be written as:
[tex]\[ \int \frac{6x + 2}{9x^2 + 6x + 26} \, dx = \int \frac{1/3 (18x + 6)}{9x^2 + 6x + 26} \, dx \][/tex]
Thus, we can now separate:
[tex]\[ = \frac{1}{3} \int \frac{18x + 6}{9x^2 + 6x + 26} \, dx \][/tex]
Rewriting this, we get a clearer form where each part is related to a derivative of the denominator:
[tex]\[ = \frac{1}{3} \int \frac{d(9x^2 + 6x + 26)}{9x^2 + 6x + 26} \][/tex]
### Step 5: Integrate Using Logarithm Rule
The integral of a function of the form [tex]\(\frac{du}{u}\)[/tex] is [tex]\(\ln|u|\)[/tex]. Therefore, we have:
[tex]\[ \int \frac{d(9x^2 + 6x + 26)}{9x^2 + 6x + 26} = \ln|9x^2 + 6x + 26| + C \][/tex]
Thus, our integral becomes:
[tex]\[ = \frac{1}{3} \ln|9x^2 + 6x + 26| + C \][/tex]
Since [tex]\(9x^2 + 6x + 26\)[/tex] is always positive for all [tex]\(x\)[/tex] (as it does not have real roots), the absolute value can be dropped:
[tex]\[ = \frac{1}{3} \ln(9x^2 + 6x + 26) + C \][/tex]
### Conclusion
Therefore, the solution to the integral [tex]\(\int \frac{6x + 2}{9x^2 + 6x + 26} \, dx\)[/tex] is:
[tex]\[ \boxed{\frac{1}{3} \ln(9x^2 + 6x + 26) + C} \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
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