Integrate the functions.
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,
⇒
⇒