Q. 194.9( 7 Votes )

# Find:

Answer :

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     Rate this question :

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