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int(cosx-sinx+1-x)/(e^(x)+sinx+x)dx=log(...

`int(cosx-sinx+1-x)/(e^(x)+sinx+x)dx=log_(e)(f(x))+g(x)+C` where C is the constant of integration and f(x) is positive. Then `f(x)+g(x)` has the value equal to

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