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Question

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


A

12log(e4xe2x+1e4x+e2x+1)+c

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B

12log(e2x+ex+1e2xex+1)+c

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C

12log(e2xe2x+1e2x+ex+1)+c

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D

12log(e4xe2x+1e4xe2x+1)+c

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Solution

The correct option is C

12log(e2xe2x+1e2x+ex+1)+c



Given I=exe4x+e2x+1dxJ=exe4x+e2x+1dx=e3xe4x+e2x+1JI=ex(e2x1)e4x+e2x+1dxLet ex=texdx=dtJI=t21t4+t2+1dt=11t2t2+1+1t2dtLet t+1t=u(11t2)dt=duJI=duu21=12logu1u+1+c=12log∣ ∣t2+1t1t2+1t+1∣ ∣+c=12loge2xex+1e2x+ex+1+c


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