Risolvere X SOSTITUZIONE.
a.
Poniamo t=3−4x⇒−14dt=dx
∫dx3−4x=
=−14∫t−12dt=
=−142t12+c=
=−12(3−4x)12+c=
=−123−4x+c
b.
Poniamo t=x+1⇒dt=dx
∫dxx+1=
=∫(x+1)−12dx=
=2t12+c=
=2t+c=
=2x+1+c
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