Integrate the functions.

Let 5x - 2 =

⇒ 5x - 2 = A(2 + 6x) + B

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

6A = 5

⇒ A =

2A + B = -2

⇒ B =

⇒ 5x - 2 = (2 + 6x) +

⇒

⇒

Now,

Let 1+2x+3x² = t

⇒ (2 + 6x)dx = dt

⇒

= log|1+2x+3x²| …(1)

And

1+2x+3x² =

⇒

⇒

⇒

⇒

…(2)

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

⇒

⇒