Let 5x + 3 =

⇒ 5x + 3 = A(2x + 4) + B

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

2A = 5

⇒ A =

4A + B = 3

⇒ B = -7

⇒ 5x + 3 =

Now,

⇒

Now, let us consider,

Let x² +4x + 10= t

⇒ (2x + 4) dx= dt

… (1)

And, Now let us consider,

⇒

⇒

…(2)

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

⇒

⇒