Find:

Let

Multiplying and dividing by x,

Let x² = t

⇒ 2x dx = dt

Now,

⇒ t² + 1 = A(t+1)² + B t(t+1) + Ct

At t = 0, we get, A = 1

At t = -1, we get, C = -2

At t = 1, we get B = 0

We can also find A, B, C by the following method,

We have,

t² + 1 = A(t+1)² + B t(t+1) + Ct

⇒ t² + 1 = A(t² + 2t + 1)² + B(t² + 1) + Ct

⇒ t² + 1 = A(t² + 2t + 1)² B(t² + 1) + Ct

⇒ t² + 1 = At² + 2At + A + Bt² + Bt + Ct

⇒ t² + 1 = (A+B)t² + (2A+B+C)t + A

On comparing coefficient of t², t, constant, we get,

A + B = 1 … (1)

2A + B + C = 0 … (2)

A = 1 … (3)

From (3), A = 1,

Put A = 1 in (1)

1 + B = 1

⇒ B = 1 – 1

⇒ B = 0

So, put A = 1, B = 0 in (2) we get,

2(1) + 0 + C = 0

⇒ C = - 2

So, A = 1, B = 0, C = -2