Resposta :
[tex] \int {(x^3-3x^2+senx)} \, dx =\int x^3-3 \int x^2+ \int senx= \\
\\
\frac{x^4}{4}-3\frac{x^3}{3}-cos x \\
\\
\frac{x^4}{4}-x^3-cosx + C[/tex]
Calcule a integral
∫(x³-3x²+senx)dx