Evaluate the integral: 37/x^3-125
To evaluate the integral ∫37/(x^3-125) dx, we can use partial fraction decomposition and the substitution method.
1. First, let's factorize the denominator. Notice that it is a difference of cubes: x^3 - 125 = (x - 5)(x^2 + 5x + 25).
2. Now, we can express the fraction as the sum of partial fractions: 37/(x^3-125) = A/(x - 5) + B(x^2 + 5x + 25)/(x^3-125).
3. Multiply both sides of the equation by (x^3-125) to eliminate the fractions: 37 = A(x^2 + 5x + 25) + B(x - 5).
4. Expand the right side: 37 = Ax^2 + 5Ax + 25A + Bx - 5B.
5. Group the terms based on powers of x: 37 = (A + B)x^2 + (5A + B)x + 25A - 5B.
6. Equate the coefficients of like powers of x: A + B = 0 (coefficient of x^2), 5A + B = 0 (coefficient of x), and 25A - 5B = 37 (constant term).
7. Solve the system of equations to find the values of A and B. From the first equation, we get A = -B. Substituting this in the second equation, we get 5(-B) + B = 0, which gives B = 0. Substituting B = 0 in the third equation, we get 25A - 5(0) = 37, which gives A = 37/25.
8. Now that we have the values of A and B, we can rewrite the original integral as ∫(37/25)/(x - 5) dx.
9. Use the substitution method: Let u = x - 5. Then, du = dx.
10. Rewrite the integral in terms of the new variable: ∫(37/25) du.
11. Integrate: (37/25)u + C, where C is the constant of integration.
12. Substitute the original variable back: (37/25)(x - 5) + C.
Therefore, the evaluated integral is (37/25)(x - 5) + C.
To evaluate the integral ∫(37/(x^3-125)) dx, we can use the method of partial fractions.
First, let's factor the denominator x^3-125 using the difference of cubes formula:
x^3-125 = (x-5)(x^2+5x+25)
Now, we can express the fraction 37/(x^3-125) in its partial fraction form:
37/(x^3-125) = A/(x-5) + (Bx + C)/(x^2+5x+25)
To find the values of A, B, and C, we need to clear the denominators by multiplying both sides of the equation by (x-5)(x^2+5x+25):
37 = A(x^2+5x+25) + (Bx + C)(x-5)
Let's expand the right side and then match the coefficients of corresponding powers of x:
37 = Ax^2 + 5Ax + 25A + Bx^2 - 5Bx + Cx - 5C
Now, let's group the terms with the same powers of x:
37 = (A + B)x^2 + (5A - 5B + C)x + (25A - 5C)
Matching the coefficients, we have the following equations:
A + B = 0 (since there's no x^2 term on the left side)
5A - 5B + C = 0 (since the coefficient of x in the left side is 0)
25A - 5C = 37 (since the constant term on the left side is 37)
Solving these equations simultaneously, we find A = -37/150, B = 37/150, and C = 37/5.
Now that we have the partial fraction decomposition:
37/(x^3-125) = -37/150/(x-5) + 37/150x + 37/5/(x^2+5x+25)
We can now integrate each term separately:
∫(-37/150/(x-5)) dx + ∫(37/150x) dx + ∫(37/5/(x^2+5x+25)) dx
The first term is a simple logarithmic function:
−(37/150)ln|x-5| + ∫(37/150x) dx + ∫(37/5/(x^2+5x+25)) dx
The second term is a straightforward linear function:
−(37/150)ln|x-5| + (37/150)x + ∫(37/5/(x^2+5x+25)) dx
To integrate the last term, we can complete the square in the denominator:
−(37/150)ln|x-5| + (37/150)x + ∫(37/5/(x^2+5x+25)) dx
−(37/150)ln|x-5| + (37/150)x + ∫(37/5/((x+5/2)^2+(15/2)^2)) dx
Now, we can make a substitution using the form u = (x+5/2):
−(37/150)ln|x-5| + (37/150)x + (37/5)(2/15)∫(1/(u^2+(15/2)^2)) du
Simplifying further:
−(37/150)ln|x-5| + (37/150)x + (74/75)∫(1/(u^2+(15/2)^2)) du
The integral in the last term can be written as:
(74/75) * arctan(u/(15/2))
So, substituting back in the original variable x:
−(37/150)ln|x-5| + (37/150)x + (74/75) * arctan((x+5/2)/(15/2))
And that is the final form of the integral.