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Ur�ete integr�l:

$$\int  {a^x \over b^x}dx$$

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$$\int  {a^x \over b^x}dx = \int{\left(a \over b\right)}^xdx=$$

Zavedeme substituci:
$$\align {\left(a \over b\right)}^x&=z \\ x (\text{ln}\ a - \text{ln}\ b) &=\text{ln}\  z\\ dx&={dz \over z(\text{ln}\ a -\text{ln}\ b)}\endalign$$

$$= \int{\left(a \over b\right)}^xdx= \int {z\ dz \over z(\text{ln}\ a -\text{ln}\ b)} = \int {dz \over \text{ln}\ a -\text{ln}\ b} = {z \over \text{ln}\ a -\text{ln}\ b} = {a^x \over b^x \text{ln}\ \left({a \over b}\right)}
$$


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