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Určete integrál:

$$\int {e^x \over e^x + e^{-x}} dx$$

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Zavedeme substituci:
$$\align e^x&=z \\ e^x\ dx&=dz \\ dx&={dz \over z}\endalign$$

$$\int {e^x \over e^x + e^{-x}} dx = \int{z\ dz \over z\left( z + {1 \over z}\right)} = \int {z\ dz \over z^2 +1}=$$

Substituujeme dále:
$$\align z^2=t \\ 2z\ dz&=dt \\ z\ dz&={dt \over 2}\endalign$$

$$=\int {dt \over 2(t+1)} = {1 \over 2} \text{ln}\ (z^2 + 1)={1 \over 2} \text{ln}\ (e^{2x} + 1)$$

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