Ответы на вопрос:
Int (1 - x + 2x^2)*sin 4x dx = int sin 4x dx - int x*sin 4x dx + 2*int x^2*sin 4x dx = = -1/4*cos 4x - |u=x, dv=sin 4x dx, du=dx, v=-1/4*cos 4x| + + |u=x^2, dv=sin 4x dx, du=2x dx, v=-1/4*cos 4x| = = -1/4*cos 4x + x/4*cos 4x - 1/4*int cos 4x dx -- x^2/2*cos 4x + int x*cos 4x dx = = -1/4*cos 4x + x/4*cos 4x - 1/16*sin 4x - x^2/2*cos 4x + + |u=x, dv=cos 4x dx, du=dx, v=1/4*sin 4x| = = -1/4*cos 4x + x/4*cos 4x - 1/16*sin 4x - x^2/2*cos 4x + + x/4*sin 4x - 1/4*int sin 4x dx = = -1/4*cos 4x + x/4*cos 4x - 1/16*sin 4x - x^2/2*cos 4x + + x/4*sin 4x + 1/16*cos 4x + c
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