Risolvere X SOSTITUZIONE.
Poniamo t=(x−1)13⇒x=t3+1⇒3t2dt=dx
∫xx−13dx=
=∫t3+1t⋅3t2dt=
=3∫t4+tdt=
=35t5+32t2+c=
=35t3⋅t2+32t2+c=
=35(x−1)(x−1)23+32(x−1)23+c
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