Spiegare i passaggi.
∫(1/(x^2 + 8·x + 16))dx=
=∫(1/(x + 4)^2)dx=###
x+4=t-----> x=t-4
dx=dt
###=∫(1/t^2)dt=
=t^(-2+1)/(-2+1)=
=-t^(-1)=-1/(x+4) + C
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