$int(x/(sqrt(2-3x^2)))dx$

A cura di: Francesco Speciale

Svolgimento:
$int(x/(sqrt(2-3x^2)))dx=-1/6int((2-3x^2)^(-1/2)(-6))dx=$
$=-1/6int((2x-3x^2)^(-1/2)D(2-3x^2))dx=-1/6((2-3x^2)^(1/2)/(1/2))+c=$
$=-1/3sqrt(2-3x^2)+c$.