Given:

Using partial differentiation:

let

⇒5x=A(x²+9)+(Bx+C)(x+1)

⇒ 5x = Ax² +9A+ Bx² +Bx+ Cx + C

⇒ 5x = 9A + C + (B+C)x + (A+B)x²

Equating the coefficients of x, x² and constant value. We get:

(a) 9A + C = 0 ⇒ C = -9A

(b) B+C = 5 ⇒ B = 5-C ⇒ B = 5-(-9A) ⇒ B = 5 + 9A

( c) A + B =0 ⇒ A = -B ⇒ A = -(5 + 9A) ⇒ 10A = -5 ⇒ A = -1/2

and C = 9/2 and B = 1/2

Put these values in equation (1)

Put x² = t ⇒ 2xdx = dt

Put the value in equ. (2)