SPiegare i passaggi.
1/(x^3 + 4·x^2) = 1/(x^2·(x + 4))
1/(x^2·(x + 4)) = a/x + b/x^2 + c/(x + 4)
1/(x^2·(x + 4)) = (x^2·(a + c) + x·(4·a + b) + 4·b)/(x^2·(x + 4))
{a + c = 0
{4·a + b = 0
{4·b = 1
quindi: [ a = - 1/16 ∧ b = 1/4 ∧ c = 1/16]
1/(x^2·(x + 4)) = - 1/(16·x) + 1/(4·x^2) + 1/(16·(x + 4))
∫(1/(x^3 + 4·x^2))dx=LN|x + 4|/16 - LN|x|/16 - 1/(4·x) + C