using definite integral find area under the curve f(x)=x^-1 + x^2, between the x-values, x=1 to x=2

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The function `f(x) = 1/x + x^2` .

The area under the curve of f(x) for values of x = 1 to x = 2 is the integral:

`int_1^2 1/x + x^2 dx`

= `(ln x)_1^2 + (x^3/3)_1^2`

= `ln 2 - ln 1 + 8/3 - 1/3`

= `ln 2 + 7/3`

**The required area is **`ln 2 + 7/3`

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