Risolvere X SOSTITUZIONE.
Poniamo t=2x+13⇒x=t3−12⇒dx=32t2dt
∫x22x+13dx=
=32∫(t3−1)24tt2dt=
=38∫(t3−1)2⋅tdt=
=38∫t7−2t4+tdt=
=38[t88−2t55+t22]+c=
=364t8−320t5+316t2+c=
=364(2x+1)83−320(2x+1)53+316(2x+1)23+c
Home
Profilo
Menu