QJEEAdv08Ma0101: JEE Advanced 2008 Mathematics - Class 12 JEEAdv08Ma0101 Solution: Let Then for an arbitrary constant , the value of equals :

Given,

$I = \int {{{{e^x}} \over {{e^{4x}} + {e^{2x}} + 1}}dx}$

$J = \int {{{{e^{ - x}}} \over {{e^{ - 4x}} + {e^{ - 2x}} + 1}}dx = \int {{{{e^{3x}}} \over {{e^{4x}} + {e^{2x}} + 1}}dx} }$

$\therefore$ $J - I = \int {{{{e^x}({e^{2x}} - 1)} \over {{e^{4x}} + {e^{2x}} + 1}}dx}$

Let ${e^x} = t \Rightarrow {e^x}dx = dt$

$\therefore$ $J - I = \int {{{{t^2} - 1} \over {{t^4} + {t^2} + 1}}dt}$

$= \int {{{1 - {1 \over {{t^2}}}} \over {{t^2} + 1 + {1 \over {{t^2}}}}}dt}$

Let, $t + {1 \over t} = u \Rightarrow \left( {1 - {1 \over {{t^2}}}} \right)dt = du$

$J - I = \int {{{du} \over {{u^2} - 1}} = {1 \over 2}\log \left| {{{u - 1} \over {u + 1}}} \right| + C}$

$= {1 \over 2}\log \left| {{{{{{t^2} + 1} \over t} - 1} \over {{{{t^2} + 1} \over t} + 1}}} \right| + C$

$= {1 \over 2}\log \left| {{{{e^{2x}} - {e^x} + 1} \over {{e^{2x}} + {e^x} + 1}}} \right| + C$