Risolvere X SOSTITUZIONE.
Poniamo t=32x⇒233dt=dx
∫14−3x2dx=
=∫14−343t2dx=
=∫14−4t2233dx=
=233∫121−t2dx=
=33∫11−t2dx=
=33arcsin(t)+=
=33arcsin(x32)+c
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