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







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