Let x + 3 =

⇒ x + 3 = A(2x-2) + B

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

2A = 1

⇒ A =

-2A + B = 3

⇒ B = 4

⇒ x + 3 =

Now,

⇒

Now, Let us consider

Let x2 – 2x – 5 = t

⇒ (2x -2)dx = dt

And, now let us consider,

⇒

⇒

…(2)

Using eq. (1) and (2), we get,

⇒

⇒