Integrate the functions.

Let 4x + 1 =

⇒ 4x + 1 = A(4x + 1) + B

⇒ 4x + 1 = 4Ax+ A+ B

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

4A = 4

⇒ A = 1

A + B = 1

⇒ B = 0

Let 2x^{2} + x – 3 =t

⇒ (4x + 1) dx = dt

⇒

2