If, fx−4x+2=2x+1,(x∈R−{1,−2}) then, ∫f(x)dx is equal to : (where C is a constant of integration

If, fx4x+2=2x+1,(xR{1,2}) then, f(x)dx is equal to : (where C is a constant of integration

  1. A

    12loge|1x|3x+C

  2. B

    12loge|1x|3x+C

  3. C

    12loge|1x|+3x+C

  4. D

    12loge|1x|+3x+C

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    Solution:

    Given, fx4x+2=2x+1

    Put, x4x+2=tx4=2t+xt

    xxt=2t+4x(1t)=2(t+2)x=2(t+2)1t

      (i) becomes f(t)=22(t+2)1t+1=4(t+2)t1+1

    or f(x)=4(x+2)x1+1                             …(ii)

    On integrating (ii), we get

    f(x)dx=4(x+2)x1dx+1dx=4x+31x1dx+x=41dx121x1dx+x=3x+1211xdx=3x+12loge|1x|+C

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