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Určete integrál:

$$\int  x \sqrt{a+x} dx$$

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Zavedeme substituci:
$$\align \sqrt{a+x}&=z \\ a+x&=z^2\\ dx&=2z\ dz \\ x&=z^2-a\endalign$$

$$\int x \sqrt{a+x} dx = \int 2(z^2-a)z^2\ dz = 2 \int (z^4-az^2)\ dz = {2 \over 5 z^5}- {2a \over 3} z^3={2 \over 15}z^3(3z^2-5a)={2 \over 15} \sqrt{(a+x)^3} (3x-2a)
$$


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