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To solve the integral [tex]\(\int \frac{dx}{9x^2 + 12x + 13}\)[/tex], we proceed through the following steps:
### Step 1: Recognize the Form
The integrand [tex]\(\frac{1}{9x^2 + 12x + 13}\)[/tex] is a rational function, and we notice that the quadratic expression in the denominator can be completed to a perfect square form.
### Step 2: Completing the Square
First, we complete the square for the quadratic expression in the denominator:
[tex]\[9x^2 + 12x + 13\][/tex]
Factor out the 9 from the terms involving [tex]\(x\)[/tex]:
[tex]\[9(x^2 + \frac{4x}{3}) + 13\][/tex]
To make the expression inside the parenthesis a perfect square, we add and subtract the square of half the coefficient of [tex]\(x\)[/tex]:
[tex]\[x^2 + \frac{4x}{3} = \left(x + \frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 \][/tex]
Thus, we rewrite the expression:
[tex]\[9 \left[\left(x + \frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\right] + 13\][/tex]
[tex]\[9 \left[\left(x + \frac{2}{3}\right)^2 - \frac{4}{9}\right] + 13\][/tex]
[tex]\[9 \left(x + \frac{2}{3}\right)^2 - 4 + 13\][/tex]
[tex]\[9 \left(x + \frac{2}{3}\right)^2 + 9\][/tex]
### Step 3: Simplify the Denominator
Now the denominator becomes:
[tex]\[9 \left[\left(x + \frac{2}{3}\right)^2 + 1\right]\][/tex]
### Step 4: Substitute and Rewrite the Integral
To simplify the integral, we use the substitution:
[tex]\[u = x + \frac{2}{3}\][/tex]
[tex]\[du = dx\][/tex]
Rewriting the integral in terms of [tex]\(u\)[/tex]:
[tex]\[\int \frac{du}{9(u^2 + 1)}\][/tex]
### Step 5: Evaluate the Integral
We recognize that the integral is now in a standard form:
[tex]\[\int \frac{du}{u^2 + 1} = \arctan(u) + C\][/tex]
Thus,
[tex]\[\int \frac{du}{9(u^2 + 1)} = \frac{1}{9}\arctan(u) + C\][/tex]
### Step 6: Substitute Back
Finally, substitute back [tex]\(u = x + \frac{2}{3}\)[/tex]:
[tex]\[ \int \frac{dx}{9x^2 + 12x + 13} = \frac{1}{9}\arctan\left(x + \frac{2}{3}\right) + C \][/tex]
So the result is:
[tex]\[ \boxed{\frac{1}{9} \arctan \left( x + \frac{2}{3} \right) + C} \][/tex]
### Step 1: Recognize the Form
The integrand [tex]\(\frac{1}{9x^2 + 12x + 13}\)[/tex] is a rational function, and we notice that the quadratic expression in the denominator can be completed to a perfect square form.
### Step 2: Completing the Square
First, we complete the square for the quadratic expression in the denominator:
[tex]\[9x^2 + 12x + 13\][/tex]
Factor out the 9 from the terms involving [tex]\(x\)[/tex]:
[tex]\[9(x^2 + \frac{4x}{3}) + 13\][/tex]
To make the expression inside the parenthesis a perfect square, we add and subtract the square of half the coefficient of [tex]\(x\)[/tex]:
[tex]\[x^2 + \frac{4x}{3} = \left(x + \frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 \][/tex]
Thus, we rewrite the expression:
[tex]\[9 \left[\left(x + \frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\right] + 13\][/tex]
[tex]\[9 \left[\left(x + \frac{2}{3}\right)^2 - \frac{4}{9}\right] + 13\][/tex]
[tex]\[9 \left(x + \frac{2}{3}\right)^2 - 4 + 13\][/tex]
[tex]\[9 \left(x + \frac{2}{3}\right)^2 + 9\][/tex]
### Step 3: Simplify the Denominator
Now the denominator becomes:
[tex]\[9 \left[\left(x + \frac{2}{3}\right)^2 + 1\right]\][/tex]
### Step 4: Substitute and Rewrite the Integral
To simplify the integral, we use the substitution:
[tex]\[u = x + \frac{2}{3}\][/tex]
[tex]\[du = dx\][/tex]
Rewriting the integral in terms of [tex]\(u\)[/tex]:
[tex]\[\int \frac{du}{9(u^2 + 1)}\][/tex]
### Step 5: Evaluate the Integral
We recognize that the integral is now in a standard form:
[tex]\[\int \frac{du}{u^2 + 1} = \arctan(u) + C\][/tex]
Thus,
[tex]\[\int \frac{du}{9(u^2 + 1)} = \frac{1}{9}\arctan(u) + C\][/tex]
### Step 6: Substitute Back
Finally, substitute back [tex]\(u = x + \frac{2}{3}\)[/tex]:
[tex]\[ \int \frac{dx}{9x^2 + 12x + 13} = \frac{1}{9}\arctan\left(x + \frac{2}{3}\right) + C \][/tex]
So the result is:
[tex]\[ \boxed{\frac{1}{9} \arctan \left( x + \frac{2}{3} \right) + C} \][/tex]
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