$\int_{0}^{1} \frac{e^{x}}{1 + e^{2 x}} d x = ?$ $\int_0^1e^x/(1+e^(2x))dx=?$

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$\int_{0}^{1} \frac{e^{x}}{1 + e^{2 x}} d x = ?$ $\int_0^1e^x/(1+e^(2x))dx=?$

please don't use LaGrange ( $λ$ $\lambda$ ), I'm only in Calculus II

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Answer 1 · 2017-09-15T20:42:34

$arctan (e) - \frac{π}{4}$ $arctan(e)-pi/4$

Explanation:

Let $u = e^{x}$ $u=e^x$ . Thus $d u = e^{x} d x$ $du=e^xdx$ . The bounds are also transformed: $x = 0 \Rightarrow u = e^{0} = 1$ $x=0=>u=e^0=1$ and $x = 1 \Rightarrow u = e^{1} = e$ $x=1=>u=e^1=e$ .

$\int_{0}^{1} \frac{e^{x}}{1 + e^{2 x}} d x = \int_{1}^{e} \frac{1}{1 + u^{2}} d u$ $int_0^1e^x/(1+e^(2x))dx=int_1^e1/(1+u^2)du$

This is the arctangent integral:

$= arctan (u) ∣_{1}^{e} = arctan (e) - arctan (1) = arctan (e) - \frac{π}{4}$ $=arctan(u)|_1^e=arctan(e)-arctan(1)=arctan(e)-pi/4$

$\approx 0.43288474$ $approx0.43288474$

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$\int_{0}^{1} \frac{e^{x}}{1 + e^{2 x}} d x = ?$ $\int_0^1e^x/(1+e^(2x))dx=?$

please don't use LaGrange ( $λ$ $\lambda$ ), I'm only in Calculus II

1 Answer

Explanation:

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∫10ex1+e2xdx=?\int_0^1e^x/(1+e^(2x))dx=?

please don't use LaGrange (λ\lambda), I'm only in Calculus II

1 Answer

Explanation:

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$\int_{0}^{1} \frac{e^{x}}{1 + e^{2 x}} d x = ?$ $\int_0^1e^x/(1+e^(2x))dx=?$

please don't use LaGrange ( $λ$ $\lambda$ ), I'm only in Calculus II