Integrate the functions.
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,
⇒
⇒