How to integrate~ [(2x-1)/(x^2-8x+15)] dx What method that can be use to evaluate?

Use integration by partial fractions.

x² - 8x + 15 = (x - 3)(x - 5)
Let (2x - 1)/(x² - 8x + 15) = [A/(x - 3)] + [B/(x - 5)]
Then A(x - 5) + B(x - 3) = 2x - 1

Put x = 3: A(3 - 5) = 2(3) - 1, then A = -5/2
Put x = 5: B(5 - 3) = 2(5) - 1, then B = 9/2
Thus, (2x - 1)/(x² - 8x + 15) = [9/2(x - 5)] - [5/2(x - 3)]

∫[(2x - 1)/(x² - 8x + 15)] dx
= ∫{[9/2(x - 5)] - [5/2(x - 3)]} dx
= ∫[9/2(x - 5)] dx - ∫[5/2(x - 3)] dx
= (9/2)ln|x - 5| - (5/2)ln|x - 3| + C

∫ (2x-1) /(x^2-8x+15) dx

2x-1 = 2x-8 + 7
∫ (2x-1) /(x^2-8x+15) dx = ∫ (2x-8) dx /(x^2-8x+15) + ∫ 7 dx /(x^2-8x+15)

Integrate ∫ (2x-8) dx/(x^2-8x+15)

Let u= x^2-8x+17
du = (2x-8) dx
∫ (2x-8) dx/(x^2-8x+15) = ∫ du/u = ln |u| = ln |x^2-8x+15| ------(1)

Integrate 7 ∫ dx/(x^2-8x+15)

x^2-8x+15 = x^2-8x+16 -1 = (x-4)^2 - 1
7 ∫ dx /(x^2-8x+15) = 7 ∫ dx /((x-4)^2 -1)

Let u=x-4
du = dx
7 ∫ dx/ ((x-4)^2 -1) = 7 ∫ du /(u^2-1) = 7 ∫ du/((u-1)(u+1))

7 /((u-1)(u+1)) = A/(u-1) + B/(u+1)
multiply both sides by (u-1)(u+1)
7 = A (u+1) + B(u-1)

Let u=1
7 = A(2) + B(0)
2A=7
A = 7/2

Let u=-1
7 = A(0) + B(-2)
-2B= 7
B = -7/2

7 /((u-1)(u+1)) = A/(u-1) + B/(u+1)
∫ 7 du /((u-1)(u+1)) = (7/2) ∫ du/(u-1) - (7/2) ∫ du/(u+1)
= (7/2) ln |u-1| - (7/2) ln |u+1|

replace u by x-4
= (7/2) ln | x-5| - (7/2) ln |x-3| --------(2)

(1)+(2) =

= ln |x^2-8x+15| + (7/2) ln | x-5| - (7/2) ln |x-3| + C