Let 6x + 7 =

⇒ 6x + 7 = A(2x - 9) + B

Now, equating the coefficients of x and constant term on both sides, we get,

2A = 6

⇒ A = 3

-9A + B = 7

⇒ B = 34

⇒ 6x + 7 = 3 (2x - 9) + 34

⇒

⇒

Now,

Let x² – 9x + 20 = t

⇒ (2x - 9)dx = dt

----------(1)

And

x² – 9x + 20 =

⇒

⇒

…(2)

Thus, from (1) and (2), we get,

⇒

⇒