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We note that x(f(x)) is the integrand above. We also note that the poles of f are z±=-1±2i. By choosing the right contour, one can show that -+f(x)dx=-2iπ(z)<0zresidueRes(f,z) and thus -+16sin(2x+π)5(x2+2x+5)2dx=πsin(2)e4