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[tex] \int\ {(x^2+2x)cos2x} \, dx [/tex]

[tex] \int\ {(x^2+2x)cos2x} \, dx\\u=x^2+2x=\ \textgreater \ du=(2x+2)dx\\dv=cos2xdx=\ \textgreater \ v=\frac{1}{2}sin2x\\ \int{(x^2+2x)cos2x} dx=\frac{(x^2+2x)sin2x}{2}-\frac{1}{2}\int sin2x(2x+2)dx\\\\\int sin2x(2x+2)dx\\u=2x+2=\ \textgreater \ du=2dx\\dv=sin2x=\ \textgreater \ v=-\frac{1}{2}cos2x\\\int sin2x(2x+2)dx=-\frac{(2x+2)cos2x}{2}+\int cos2xdx=\\-\frac{(2x+2)cos2x}{2}+\frac{1}{2}sin2x\\\\ \int\ {(x^2+2x)cos2x} \, dx=\frac{(x^2+2x)sin2x}{2}-\frac{1}{2}(-\frac{(2x+2)cos2x}{2}+\frac{1}{2}sin2x)=\\=\frac{(x^2+2x)sin2x}{2}+\frac{(2x+2)cos2x}{4}-\frac{sin2x}{4}+C=[/tex]
[tex]=\frac{(2x^2+4x)sin2x-sin2x}{4}+\frac{(2x+2)cos2x}{4}+C=\\=\frac{sin2x(2x^2+4x-1)}{4}+\frac{(2x+2)cos2x}{4}+C=\\=\frac{1}{4}(sin2x(2x^2+4x-1)+2cos2x(x+1))[/tex]