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Question

If dxx2(xn+1)(n1)n=[f(x)]1n+C then f(x) is


A

(1+xn)

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B

1+xn

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C

xn+xn

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D

None of these

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Solution

The correct option is B

1+xn


dxx2(xn+1)(n1)n=dxx2.xn1(1+1xn)(n1)n
=dxxn+1(1+xn)(n1)n
Put 1+xn=t
nxn1dx=dtordxxn+1=dtn
=dxxn+1(1+xn)(n1)n=1ndtt((n1)n)
=1nt1n1dt=1n.t1n1+11n1+1+c=t1n +c
=(1+xn)1n +c


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