Spiega il ragionamento.
∫(1/(x + 2·√x)dx=
pongo:
√x = t---> x = t^2---> dx=2·t dt
=∫(1/(t^2 + 2·t)·2·t)dt =
=2·∫(1/(t + 2))dt=
=2·LN|t + 2|=
=2·LN|√x + 2| + C
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