evaluate the integral:
integral of 13/(x^3 - 27) dx
use partial fractions to get
13/27 (1/(x-3) - (x+6)/(x^2+3x+9))
13/27 (1/(x-3) - 1/2 (2x+3)/(x^2+3x+9) - (9/2)/(x^2+3x+9))
13/27 (1/(x-3) - 1/2 (2x+3)/(x^2+3x+9) - (9/2)/(x+3/2)^2 + 27/4)
Now you can use substitutions
u = x-3
v = x^2+3x+9
w = x + 3/2
to come up with
13/27 (du/u - 1/2 dv/v - 9/2 / (w^2+(√27/2)^2))
Then using the trig substitution for
∫ 1/(w^2+a^2) = 1/a arctan(w/a)
everything falls right out.
To evaluate the integral of 13/(x^3 - 27) dx, we can first factor the denominator as a difference of cubes:
x^3 - 27 = (x - 3)(x^2 + 3x + 9)
The integral can then be written as:
∫(13/[ (x - 3)(x^2 + 3x + 9) ]) dx
We can use partial fraction decomposition to simplify this integral further. Let's decompose the integrand:
13/[ (x - 3)(x^2 + 3x + 9) ] = A/(x - 3) + (Bx + C)/(x^2 + 3x + 9)
To find the values of A, B, and C, we need to equate the numerator of the integrand to the right-hand side:
13 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)
Expanding and gathering like terms, we get:
13 = (A + B)x^2 + (3A - 3B + C)x + (9A - 3C)
Equating the coefficients of matching powers of x, we get the following equations:
A + B = 0 (coefficient of x^2 term)
3A - 3B + C = 0 (coefficient of x term)
9A - 3C = 13 (constant term)
From the first equation, we can solve for B in terms of A:
B = -A
Substituting this into the second equation, we have:
3A - 3(-A) + C = 0
6A + C = 0
C = -6A
Substituting the value of C into the third equation, we have:
9A - 3(-6A) = 13
9A + 18A = 13
27A = 13
A = 13/27
Using this value of A, we can find B and C:
B = -A = -13/27
C = -6A = -6(13/27)
Now that we have the values of A, B, and C, we can rewrite the integral in terms of partial fractions:
∫(13/[ (x - 3)(x^2 + 3x + 9) ]) dx = ∫(A/(x - 3) + (Bx + C)/(x^2 + 3x + 9)) dx
= ∫(13/27)/(x - 3) dx - 13/(27∫(x^2 + 3x + 9)/(x^2 + 3x + 9)) dx
= 13/27 ∫(1/(x - 3)) dx - 13/27 ∫(1) dx
The integral of 1/(x - 3) is ln|x - 3|, and the integral of 1 is x. Therefore, the integral simplifies to:
= 13/27 ln|x - 3| - 13/27 x + C
where C is the constant of integration.
To evaluate the given integral of 13/(x^3 - 27) dx, we first need to factor the denominator (x^3 - 27) so that we can simplify the integrand. The denominator is a difference of cubes and can be factored as follows:
x^3 - 27 = (x - 3)(x^2 + 3x + 9)
Now, we can rewrite the integral using the factored denominator:
∫(13 / (x - 3)(x^2 + 3x + 9)) dx
The next step is to separate the integrand into partial fractions. We're looking for constants A, B, and C, such that:
13 / ((x - 3)(x^2 + 3x + 9)) = A / (x - 3) + (Bx + C) / (x^2 + 3x + 9)
To determine A, B, and C, we need to find a common denominator. Multiply the entire equation by (x - 3)(x^2 + 3x + 9):
13 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)
Expanding the right-hand side:
13 = A(x^2 + 3x + 9) + Bx^2 - 3Bx + Cx - 3C
Now, equating coefficients of like terms on both sides, we get the following system of equations:
For the constant terms:
0 = 9A - 3C (1)
For the x terms:
0 = 3A + C - 3B (2)
For the x^2 terms:
0 = A + B (3)
From equation (1), we find that 9A = 3C, which simplifies to 3A = C.
Substituting this into equation (2), we have:
0 = 3A + 3A - 3B
Simplifying further:
0 = 6A - 3B
3B = 6A
Lastly, using equation (3), we can substitute B = 2A:
3B = 6A
3(2A) = 6A
6A = 6A
Since the coefficient of A on both sides of the equation is the same, we can set any value for A, such as A = 1.
Therefore, B = 2A = 2, and C = 3A = 3.
Now that we have determined the values of A, B, and C, we can rewrite the integral as follows:
∫(13 / ((x - 3)(x^2 + 3x + 9))) dx = ∫(A / (x - 3) + (Bx + C) / (x^2 + 3x + 9)) dx
Substituting the values:
∫(13 / ((x - 3)(x^2 + 3x + 9))) dx = ∫(1 / (x - 3) + (2x + 3) / (x^2 + 3x + 9)) dx
Since we have separated the integrand into two fractions, we can now evaluate each term separately.
For the first term ∫(1 / (x - 3)) dx, we can use the property of natural logarithm:
∫(1 / (x - 3)) dx = ln |x - 3| + C1
For the second term ∫((2x + 3) / (x^2 + 3x + 9)) dx, we can perform a u-substitution. Let u = x^2 + 3x + 9, then du = (2x + 3) dx:
∫((2x + 3) / (x^2 + 3x + 9)) dx = ∫(1 / u) du = ln |u| + C2
Substituting u back in terms of x:
ln |u| + C2 = ln |x^2 + 3x + 9| + C2
Putting it all together, the solution to the integral is:
∫(13 / (x^3 - 27)) dx = ln |x - 3| + ln |x^2 + 3x + 9| + C
Note: C1 and C2 are constants of integration, and C is the overall constant of integration.