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,




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