1.

If `I=int(e^x)/(e^(4x)+e^(2e)+1) dx. J=int(e^(-x))/(e^(-4x)+e^(-2x)+1) dx.` Then for an arbitrary constant c, the value of `J-I` equal toA. `(1)/(2)log|(e^(4x)-e^(2x)+1)/(e^(4x)+e^(2x)+1)|+c`B. `(1)/(2)log|(e^(2x)+e^(x)+1)/(e^(2x)-e^(x)+1)|+c`C. `(1)/(2)log|(e^(2x)-e^(x)+1)/(e^(2x)+e^(x)+1)|+c`D. `(1)/(2)log|(e^(4x)+e^(2x)+1)/(e^(4x)-e^(2x)+1)|+c`

Answer» Correct Answer - C
Since, `I = int(e^(x))/(e^(4x)+e^(2x)+1)dx and J = int (e^(3x))/(1+e^(2x)+e^(4x))dx`
`therefore J - I = int((e^(3x)-e^(x)))/(1+e^(2x) + e^(4x))dx`
Put `e^(x)=u rArr e^(x)dx=du`
`therefore J-I = int((u^(2)-1))/(1+u^(2)+u^(4))du = int((1-(1)/(u^(2))))/(1+(1)/(u^(2))+u^(2))du`
`= int ((1-(1)/(u^(2))))/((u+(1)/(u))^(2)-1)du`
`Put" "u+(1)/(u)=t`
`rArr" "(1-(1)/(u^(2)))du = dt`
`" "=int(dt)/(t^(2)-1)=(1)/(2)log|(t-1)/(t+1)|+c`
`" "=(1)/(2)log|(u^(2)-u+1)/(u^(2)+u+1)|+c`
`" "=(1)/(2)log|(e^(2x)-e^(x)+1)/(e^(2x)+e^(x)+1)|+c`


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