Integrale

∫ x^2 * ln(x + 1) dx;

f(x) = ln(x + 1);

f'(x) = 1/(1 + x);

g'(x) = x^2;

g(x) = x^3 / 3;

∫ f(x) * g'(x) dx = f(x) * g(x) - ∫ f'(x) * g(x) dx; (integrazione per parti).

∫ x^2 * ln(x + 1) dx = ln(x + 1) * x^3 /3 - ∫[1/(x + 1)  * x^3 / 3 ] dx,

ln(x + 1) * x^3 /3 -  1/3 * ∫[x^3/(x + 1)] dx;

x^3/(x+1) =  (x^3 + 1 - 1) / (x + 1) = (x^3 + 1) /(x + 1) - 1 / (x + 1);

= (x + 1) * (x2 - x + 1) / (x+1) - 1 / (x + 1) = (x^2 - x + 1) - 1/(x + 1);

ln(x + 1) * x^3 /3 -  1/3 * ∫[x^3/(x + 1)] dx =

ln(x + 1) * x^3 /3 -  1/3 * ∫[(x^2 - x + 1) - 1/(x + 1)] dx =

[1/3 * ln(x + 1) * x^3]  -  1/3 * ∫[(x^2 - x + 1) dx - 1/3 *∫[1/(x + 1)] dx =

ln(x + 1) * x^3 /3 -  1/3 * (x^3/3 - x^2/2 + x) - [1/3 * ln(x + 1)] =

1/3 ln(x + 1) * (x^3 - 1) - x^3 + x^2 /6 - x /3  + C

Ciao @mate23

non so se ho fatto bene i calcoli...