Answer:
17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C
Explanation:
∫ 17 / (x³ − 125) dx
= 17 ∫ 1 / (x³ − 125) dx
= 17 ∫ 1 / ((x − 5) (x² + 5x + 25)) dx
Use partial fraction decomposition:
= 17 ∫ [ A / (x − 5) + (Bx + C) / (x² + 5x + 25) ] dx
Use common denominator to find the missing coefficients.
A (x² + 5x + 25) + (Bx + C) (x − 5) = 1
Ax² + 5Ax + 25A + Bx² − 5Bx + Cx − 5C = 1
(A + B) x² + (5A − 5B + C) x + 25A − 5C = 1
Match the coefficients and solve the system of equations.
A + B = 0
5A − 5B + C = 0
25A − 5C = 1
A = 1/75
B = -1/75
C = -2/15
So the integral is:
= 17 ∫ [ 1/75 / (x − 5) + (-1/75 x − 2/15) / (x² + 5x + 25) ] dx
Simplify:
= 17/75 ∫ [ 1 / (x − 5) − (x + 10) / (x² + 5x + 25) ] dx
Factor ½ from the numerator of the second fraction:
= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 20) / (x² + 5x + 25) ] dx
Split the fraction:
= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − ½ (15) / (x² + 5x + 25) ] dx
Multiply the last fraction by 4/4:
= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 30 / (4x² + 20x + 100) ] dx
Complete the square:
= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 15 / ((2x + 5)² + 75) ] dx
Split the integral:
= 17/75 ∫ 1 / (x − 5) dx − 17/150 ∫ (2x + 5) / (x² + 5x + 25) dx − 17/5 ∫ 1 / ((2x + 5)² + 75) dx
The first integral is:
∫ 1 / (x − 5) dx = ln│x − 5│
The second integral is:
∫ (2x + 5) / (x² + 5x + 25) dx = ln(x² + 5x + 25)
The third integral is:
∫ 1 / ((2x + 5)² + 75) dx = 1/√75 tan⁻¹((2x + 5) / √75)
Plug in:
= 17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C