Answer :

Let

Multiplying and dividing by x,

Let x^{2} = t

⇒ 2x dx = dt

Now,

⇒ t^{2} + 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,

t^{2} + 1 = A(t+1)^{2} + B t(t+1) + Ct

⇒ t^{2} + 1 = A(t^{2} + 2t + 1)^{2} + B(t^{2} + 1) + Ct

⇒ t^{2} + 1 = A(t^{2} + 2t + 1)^{2} B(t^{2} + 1) + Ct

⇒ t^{2} + 1 = At^{2} + 2At + A + Bt^{2} + Bt + Ct

⇒ t^{2} + 1 = (A+B)t^{2} + (2A+B+C)t + A

On comparing coefficient of t^{2}, 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

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