`int_1^2(x^2+1)/(x^3+3x)dx`

Put `u = x^3+ 3x`, then `du = (3x^2+ 3) = 3(x^2+ 1)\ dx`.

`int_1^2(x^2+1)/(x^3+3x)dx=1/3int_(x=1)^(x=2)(du)/u`

`=1/3[ln|u|]_(x=1)^(x=2)`

`=1/3[ln|x^3+3x|]_1^2`

`=1/3[ln(14)-ln(4)]`

`=0.418`

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