Integrate the functions.
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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,
⇒ 
⇒ 