Supposing that you need to integrate the expression, you need to use reminder theorem to write a simpler form of the fraction such that:

`(x^3 + 3x)/(x^2 + 1) = x + 2x/(x^2 + 1) `

Integrating both sides yields:

`int (x^3 + 3x)/(x^2 + 1)dx = int x dx+ int 2x/(x^2 + 1)dx`

You need to solve the integral `int 2x/(x^2 + 1) dx` using substitution, as the problem has indicated, such that:

`x^2 + 1 = y`

Differentiating both sides yields:

`2x dx = dy`

`int 2x/(x^2 + 1) dx = int (dy)/y = ln |y| + c`

`int 2x/(x^2 + 1) dx = ln (x^2 + 1) + c`

Hence, evaluating the integral of the fraction `(x^3 + 3x)/(x^2 + 1)` yields:

`int (x^3 + 3x)/(x^2 + 1) dx = x^2 +ln (x^2 + 1) + c`

**Hence, evaluating the integral of the given function using substitution yields `int (x^3 + 3x)/(x^2 + 1) dx = x^2 + ln (x^2 + 1) + c.` **