Risolvere X SOSTITUZIONE.
Poniamo t=x+1⇒x=t−1⇒dx=dt
∫x2x+14dx=
=∫(t−1)2t4dx=
=∫(t2−2t+1)t−14dt=
=∫(t74−2t34+t−14)dt=
=411t114−247t74+43t34+c=
=411(x+1)114−87(x+1)74+43(x+1)34+c=
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