Risolvere X SOSTITUZIONE.
Poniamo t=1+x⇒x=t2−1⇒dx=2tdt
∫x1+xdx=
=∫(t2−1)t2tdt=
=2∫t4−t2dt=
=2[t55−t33]+c=
=2[(1+x)21+x5−(1+x)1+x3]+c=
=215(1+x)1+x(3(1+x)−5)+c=
=215(1+x)(3x−2)1+x+c
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