Risolvere X SOSTITUZIONE.
a.
Poniamo t=3x+2⇒dt=3dx⇒13dt=dx
∫3x+2dx=13∫t12dt=
=1323t32+c=
=29t⋅t+c=
=29(3x+2)3x+2+c
b.
Poniamo t=4−5x⇒dt=−5dx⇒−15dt=dx
∫3x+2dx=
=−15∫t13dt
=−1534t43+c=
=−320t⋅t13+c=
=−320(4−5x)4−5x3+c
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