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$\int _{-1} ^ 1 \large\frac{\cos h x }{1+e^{2x}}$$dx$ is equal to :

\[\begin {array} {1 1} (a)\;0 & \quad (b)\;1 \\ (c)\;\frac{e^2-1}{2e} & \quad (d)\;\frac{e^2+2}{2e} \end {array}\]

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1 Answer

(c) $\large\frac{e^2-1}{2e}$
answered Feb 27, 2014 by meena.p
 
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