Question
Let $$I = \int {\frac{{{e^x}}}{{{e^{4x}} + {e^{2x}} + 1}}dx,\,J = \int {\frac{{{e^{ - \,x}}}}{{{e^{ - \,4x}} + {e^{ - \,2x}} + 1}}dx.} } $$
Then for an arbitrary constant $$C,$$ the value of $$J-I$$ equals-
A.
$$\frac{1}{2}\log \left( {\frac{{{e^{4x}} - {e^{2x}} + 1}}{{{e^{4x}} + {e^{2x}} + 1}}} \right) + C$$
B.
$$\frac{1}{2}\log \left( {\frac{{{e^{2x}} + {e^x} + 1}}{{{e^{2x}} - {e^x} + 1}}} \right) + C$$
C.
$$\frac{1}{2}\log \left( {\frac{{{e^{2x}} - {e^x} + 1}}{{{e^{2x}} + {e^x} + 1}}} \right) + C$$
D.
$$\frac{1}{2}\log \left( {\frac{{{e^{4x}} + {e^{2x}} + 1}}{{{e^{4x}} - {e^{2x}} + 1}}} \right) + C$$
Answer :
$$\frac{1}{2}\log \left( {\frac{{{e^{2x}} - {e^x} + 1}}{{{e^{2x}} + {e^x} + 1}}} \right) + C$$
Solution :
$$\eqalign{
& {\text{Given}} \cr
& I = \int {\frac{{{e^x}}}{{{e^{4x}} + {e^{2x}} + 1}}dx,\,\,\,\,} \cr
& J = \int {\frac{{{e^{ - \,x}}}}{{{e^{ - \,4x}} + {e^{ - \,2x}} + 1}}dx} = \int {\frac{{{e^{3x}}}}{{{e^{4x}} + {e^{2x}} + 1}}dx} \cr
& \therefore J - I = \int {\frac{{{e^x}\left( {{e^{2x}} - 1} \right)}}{{{e^{4x}} + {e^{2x}} + 1}}dx} \cr
& {\text{Let }}{e^x} = t\,\, \Rightarrow {e^x}\,dx = dt \cr
& \therefore J - I\int {\frac{{{t^2} - 1}}{{{t^4} + {t^2} + 1}}} dt = \int {\frac{{1 - \frac{1}{{{t^2}}}}}{{{t^2} + 1 + \frac{1}{{{t^2}}}}}dt} \cr
& {\text{Let }}t - \frac{1}{t} = u\,\, \Rightarrow \left( {1 - \frac{1}{{{t^2}}}} \right)dt = du \cr
& \therefore J - I = \int {\frac{{du}}{{{u^2} - 1}} = \frac{1}{2}\log \left| {\frac{{u - 1}}{{u + 1}}} \right| + C} \cr
& = \frac{1}{2}\log \left| {\frac{{\frac{{{t^2} + 1}}{t} - 1}}{{\frac{{{t^2} + 1}}{t} + 1}}} \right| + C \cr
& = \frac{1}{2}\log \left| {\frac{{{e^{2x}} - {e^x} + 1}}{{{e^{2x}} + {e^x} + 1}}} \right| + C \cr} $$