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To solve the integral [tex]\(\int \frac{x^2 \, dx}{x^6 - 9}\)[/tex], let's work through it step by step.
First, observe that the denominator can be factored. Notice that:
[tex]\[ x^6 - 9 = (x^3)^2 - 3^2 \][/tex]
This is a difference of squares which can be factored as:
[tex]\[ x^6 - 9 = (x^3 - 3)(x^3 + 3) \][/tex]
Thus, the integral becomes:
[tex]\[ \int \frac{x^2 \, dx}{(x^3 - 3)(x^3 + 3)} \][/tex]
To proceed, we can use partial fraction decomposition to rewrite the integrand in a more manageable form:
[tex]\[ \frac{x^2}{(x^3 - 3)(x^3 + 3)} = \frac{A}{x^3 - 3} + \frac{B}{x^3 + 3} \][/tex]
To determine the constants [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we equate:
[tex]\[ x^2 = A(x^3 + 3) + B(x^3 - 3) \][/tex]
Setting the coefficients equal on both sides, we need [tex]\(A\)[/tex] and [tex]\(B\)[/tex] such that:
[tex]\[ x^2 = A x^3 + 3A + B x^3 - 3B \][/tex]
By collecting like terms, we get:
[tex]\[ x^2 = (A + B)x^3 + 3(A - B) \][/tex]
Since there is no [tex]\(x^3\)[/tex] term on the left-hand side, the coefficients of [tex]\(x^3\)[/tex] on the right-hand side must sum to zero:
[tex]\[ A + B = 0 \][/tex]
Moreover, the constant term on the right-hand side must match the [tex]\(x^2\)[/tex] term on the left-hand side:
[tex]\[ 3(A - B) = x^2 \][/tex]
From [tex]\(A + B = 0\)[/tex], we get [tex]\(B = -A\)[/tex]. Substituting this back into the equation for the constant term, we have:
[tex]\[ 3(A - (-A)) = x^2 \][/tex]
[tex]\[ 3(2A) = x^2 \][/tex]
[tex]\[ 6A = x^2 \][/tex]
[tex]\[ A = \frac{x^2}{6} \][/tex]
[tex]\[ B = -A = -\frac{x^2}{6} \][/tex]
Thus the partial fractions are:
[tex]\[ \frac{x^2}{(x^3 - 3)(x^3 + 3)} = \frac{\frac{x^2}{6}}{x^3 - 3} - \frac{\frac{x^2}{6}}{x^3 + 3} \][/tex]
Simplifying the integral, we have:
[tex]\[ \int \left( \frac{1}{6} \cdot \frac{x^2}{x^3 - 3} - \frac{1}{6} \cdot \frac{x^2}{x^3 + 3} \right) \, dx \][/tex]
Let us focus on integrating each term separately:
1. [tex]\(\frac{1}{6} \int \frac{x^2 \, dx}{x^3 - 3} \)[/tex]
2. [tex]\(-\frac{1}{6} \int \frac{x^2 \, dx}{x^3 + 3} \)[/tex]
For both integrals, we can use the substitution method. Let [tex]\(u = x^3 - 3\)[/tex], then [tex]\(du = 3x^2 \, dx\)[/tex] or [tex]\(\frac{du}{3} = x^2 \, dx\)[/tex]. Accordingly:
[tex]\[ \int \frac{x^2 \, dx}{x^3 - 3} = \int \frac{1}{3} \frac{du}{u} = \frac{1}{3} \ln |u| + C \][/tex]
Substituting back [tex]\(u = x^3 - 3\)[/tex], we get:
[tex]\[ \frac{1}{3} \ln |x^3 - 3| + C \][/tex]
Similarly, for the second integral, let [tex]\(v = x^3 + 3\)[/tex], then [tex]\(dv = 3x^2 \, dx\)[/tex] or [tex]\(\frac{dv}{3} = x^2 \, dx\)[/tex]:
[tex]\[ \int \frac{x^2 \, dx}{x^3 + 3} = \int \frac{1}{3} \frac{dv}{v} = \frac{1}{3} \ln |v| + C \][/tex]
Substituting back [tex]\(v = x^3 + 3\)[/tex], we get:
[tex]\[ \frac{1}{3} \ln |x^3 + 3| + C \][/tex]
Now, combining the two parts:
[tex]\[ \int \frac{x^2 \, dx}{x^6 - 9} = \frac{1}{6} \left( \frac{1}{3} \ln |x^3 - 3| \right) - \frac{1}{6} \left( \frac{1}{3} \ln |x^3 + 3| \right) \][/tex]
So, the final answer is:
[tex]\[ \frac{\ln |x^3 - 3|}{18} - \frac{\ln |x^3 + 3|}{18} + C \][/tex]
Therefore, the integral evaluates to:
[tex]\[ \int \frac{x^2 \, dx}{x^6 - 9} = \frac{\ln(x^3 - 3)}{18} - \frac{\ln(x^3 + 3)}{18} + C \][/tex]
Here, [tex]\(C\)[/tex] is the constant of integration.
First, observe that the denominator can be factored. Notice that:
[tex]\[ x^6 - 9 = (x^3)^2 - 3^2 \][/tex]
This is a difference of squares which can be factored as:
[tex]\[ x^6 - 9 = (x^3 - 3)(x^3 + 3) \][/tex]
Thus, the integral becomes:
[tex]\[ \int \frac{x^2 \, dx}{(x^3 - 3)(x^3 + 3)} \][/tex]
To proceed, we can use partial fraction decomposition to rewrite the integrand in a more manageable form:
[tex]\[ \frac{x^2}{(x^3 - 3)(x^3 + 3)} = \frac{A}{x^3 - 3} + \frac{B}{x^3 + 3} \][/tex]
To determine the constants [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we equate:
[tex]\[ x^2 = A(x^3 + 3) + B(x^3 - 3) \][/tex]
Setting the coefficients equal on both sides, we need [tex]\(A\)[/tex] and [tex]\(B\)[/tex] such that:
[tex]\[ x^2 = A x^3 + 3A + B x^3 - 3B \][/tex]
By collecting like terms, we get:
[tex]\[ x^2 = (A + B)x^3 + 3(A - B) \][/tex]
Since there is no [tex]\(x^3\)[/tex] term on the left-hand side, the coefficients of [tex]\(x^3\)[/tex] on the right-hand side must sum to zero:
[tex]\[ A + B = 0 \][/tex]
Moreover, the constant term on the right-hand side must match the [tex]\(x^2\)[/tex] term on the left-hand side:
[tex]\[ 3(A - B) = x^2 \][/tex]
From [tex]\(A + B = 0\)[/tex], we get [tex]\(B = -A\)[/tex]. Substituting this back into the equation for the constant term, we have:
[tex]\[ 3(A - (-A)) = x^2 \][/tex]
[tex]\[ 3(2A) = x^2 \][/tex]
[tex]\[ 6A = x^2 \][/tex]
[tex]\[ A = \frac{x^2}{6} \][/tex]
[tex]\[ B = -A = -\frac{x^2}{6} \][/tex]
Thus the partial fractions are:
[tex]\[ \frac{x^2}{(x^3 - 3)(x^3 + 3)} = \frac{\frac{x^2}{6}}{x^3 - 3} - \frac{\frac{x^2}{6}}{x^3 + 3} \][/tex]
Simplifying the integral, we have:
[tex]\[ \int \left( \frac{1}{6} \cdot \frac{x^2}{x^3 - 3} - \frac{1}{6} \cdot \frac{x^2}{x^3 + 3} \right) \, dx \][/tex]
Let us focus on integrating each term separately:
1. [tex]\(\frac{1}{6} \int \frac{x^2 \, dx}{x^3 - 3} \)[/tex]
2. [tex]\(-\frac{1}{6} \int \frac{x^2 \, dx}{x^3 + 3} \)[/tex]
For both integrals, we can use the substitution method. Let [tex]\(u = x^3 - 3\)[/tex], then [tex]\(du = 3x^2 \, dx\)[/tex] or [tex]\(\frac{du}{3} = x^2 \, dx\)[/tex]. Accordingly:
[tex]\[ \int \frac{x^2 \, dx}{x^3 - 3} = \int \frac{1}{3} \frac{du}{u} = \frac{1}{3} \ln |u| + C \][/tex]
Substituting back [tex]\(u = x^3 - 3\)[/tex], we get:
[tex]\[ \frac{1}{3} \ln |x^3 - 3| + C \][/tex]
Similarly, for the second integral, let [tex]\(v = x^3 + 3\)[/tex], then [tex]\(dv = 3x^2 \, dx\)[/tex] or [tex]\(\frac{dv}{3} = x^2 \, dx\)[/tex]:
[tex]\[ \int \frac{x^2 \, dx}{x^3 + 3} = \int \frac{1}{3} \frac{dv}{v} = \frac{1}{3} \ln |v| + C \][/tex]
Substituting back [tex]\(v = x^3 + 3\)[/tex], we get:
[tex]\[ \frac{1}{3} \ln |x^3 + 3| + C \][/tex]
Now, combining the two parts:
[tex]\[ \int \frac{x^2 \, dx}{x^6 - 9} = \frac{1}{6} \left( \frac{1}{3} \ln |x^3 - 3| \right) - \frac{1}{6} \left( \frac{1}{3} \ln |x^3 + 3| \right) \][/tex]
So, the final answer is:
[tex]\[ \frac{\ln |x^3 - 3|}{18} - \frac{\ln |x^3 + 3|}{18} + C \][/tex]
Therefore, the integral evaluates to:
[tex]\[ \int \frac{x^2 \, dx}{x^6 - 9} = \frac{\ln(x^3 - 3)}{18} - \frac{\ln(x^3 + 3)}{18} + C \][/tex]
Here, [tex]\(C\)[/tex] is the constant of integration.
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