Let I=∫exe4x+e2x+1dx,J=∫e−xe−4x+e−2x+1dx, Then, for an arbitrary constant C, the value of J-I equals

Let I=exe4x+e2x+1dx,J=exe4x+e2x+1dx, Then, for an arbitrary constant C, the value of J-I equals

  1. A

    12loge4xe2x+1e4x+e2x+1+C

  2. B

    12loge2x+ex+1e2xex+1+C

  3. C

    12loge2xex+1e2x+ex+1+C

  4. D

    12loge4x+e2x+1e4xe2x+1+C

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    Solution:

    We have,

       JI=e3xe4x2x+e2x+1exe4x+e2x+1dx  JI=e2x1e4x+e2x+1dex

     JI=t21t4+t2+1dt, where t=ex

     JI=1t+1t21dt1t JI=12logt+1t1t+tt+1+C

     JI=12logt2t+1t2+t+1+C=12loge2xex+1e2x+ex+1+C

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