If ∫ex−e−xe2x+e−2xdx=A lnex+e−x+aex+e−x−a+c, then (A, a)=

∫ex-e-xe2x+e-2x =∫ex-e-x(ex+e-x)-2dx put ex+e-x=t (ex-e-x)dx=dt I=∫dtt2-2 =122log t-2t+2+c =122log ex+e-x-2ex+e-x+2+c ∴ (A, a)=122, -2

If ∫ex−e−xe2x+e−2xdx=A lnex+e−x+aex+e−x−a+c, then (A, a)=

∫ex-e-xe2x+e-2x =∫ex-e-x(ex+e-x)-2dx put ex+e-x=t (ex-e-x)dx=dt I=∫dtt2-2 =122log t-2t+2+c =122log ex+e-x-2ex+e-x+2+c ∴ (A, a)=122, -2

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If $\int \frac{e^{x} - e^{- x}}{e^{2 x} + e^{- 2 x}} d x = A \ln |\frac{e^{x} + e^{- x} + a}{e^{x} + e^{- x} - a}| + c, t h e n (A, a) =$

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$(\frac{1}{2 \sqrt{2}}, \sqrt{2})$

$(\frac{1}{\sqrt{2}}, 2 \sqrt{2})$

$(\frac{1}{\sqrt{2}}, - 2 \sqrt{2})$

$(\frac{1}{2 \sqrt{2}}, - \sqrt{2})$

answer is D.

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Detailed Solution

$\int \frac{e^{x} - e^{- x}}{e^{2 x} + e^{- 2 x}} = \int \frac{e^{x} - e^{- x}}{(e^{x} + e^{- x}) - 2} dx put e^{x} + e^{- x} = t (e^{x} - e^{- x}) dx = dt I = \int \frac{dt}{t^{2} - 2} = \frac{1}{2 \sqrt{2}} \log |\frac{t - \sqrt{2}}{t + \sqrt{2}}| + c = \frac{1}{2 \sqrt{2}} \log |\frac{e^{x} + e^{- x} - \sqrt{2}}{e^{x} + e^{- x} + \sqrt{2}}| + c ∴ (A, a) = (\frac{1}{2 \sqrt{2}}, - \sqrt{2})$