Let I =∫exe4x+e2x+1dx.J=∫e−xe−4x+e−2x+1dx,Then, for an arbitrary constant c, the value of J-I equals
12log(e2x−e2x+1e2x+ex+1)+c
Given I=∫exe4x+e2x+1dxJ=∫e−xe−4x+e−2x+1dx=∫e3xe4x+e2x+1∴J−I=∫ex(e2x−1)e4x+e2x+1dxLet ex=t⇒exdx=dt∴J−I=∫t2−1t4+t2+1dt=∫1−1t2t2+1+1t2dtLet t+1t=u⇒(1−1t2)dt=du∴J−I=∫duu2−1=12log∣∣u−1u+1∣∣+c=12log∣∣
∣∣t2+1t−1t2+1t+1∣∣
∣∣+c=12log∣∣e2x−ex+1e2x+ex+1∣∣+c