I = ∫_1^0dx/e^x + e^ x = ∫_1^0e^x/1+e^2xdx Put e^2 = t ⇒ e^2dx=dt ∴ I = ∫_1^edt/1 + t^2=[tan^ 1t]_1^e = tan^ 1e tan^ 11 = tan^ 1e π/4 ...

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Integration Using Substitution

∫10 dx / e x+...

Question

$\int_{1}^{0} \frac{d x}{e^{x} + e^{- x}}$

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Solution

$I = \int_{1}^{0} \frac{d x}{e^{x} + e^{- x}} = \int_{1}^{0} \frac{e^{x}}{1 + e^{2 x}} d x P u t e^{2} = t \Rightarrow e^{2} d x = d t ∴ I = \int_{1}^{e} \frac{d t}{1 + t^{2}} = [t a n^{- 1} t]_{1}^{e} = t a n^{- 1} e - t a n^{- 1} 1 = t a n^{- 1} e - \frac{π}{4}$

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