∫$$01$$ $$dxex+e$$−xdx is equal to

Correct option is 2tan−1⁡e−π4

Questions

$\int_{0}^{1} \frac{dx}{e^{x} + e^{- x}} dx$ is equal to

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π4

tan−1⁡e−π4

tan−1⁡e

π4tan−1⁡e

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detailed solution

Correct option is B

∫01 dxex+e−x=∫01 exe2x+1dx Let t=ex⇒dt=exdx =∫1e dtt2+1=tan−1⁡t1e =tan−1⁡e−tan−1⁡(1)=tan−1⁡e−π4

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∫01 dxex+e−xdx is equal to

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$\int_{0}^{1} \frac{dx}{e^{x} + e^{- x}} dx$ is equal to