You need to use partial fraction decomposition such that:

`(5x^3+18x^2+43x+13)/(x^2(x^2+4x+13)) = A/x + B/x^2 + (Cx + D)/(x^2+4x+13))`

`5x^3+18x^2+43x+13 = Ax^3 + 4Ax^2 + 13Ax + Bx^2 + 4Bx + ` `13B + Cx^3 + Dx^2 `

`5x^3+18x^2+43x+13 = x^3(A+C) + x^2(4A + B + D) + x(13A + 4B) + 13B`

Equating coefficients of like powers yields:

`13B = 13 => B = ` 1

`13A + 4B = 43 => 13A = 43 - 4 => 13A = 39 => A = 3`

`4A + B + D =18 => D = 18 - 12 - 1 => D = 5`

`A+C = 5 => C = 5-3 => C = 2 `

`(5x^3+18x^2+43x+13)/(x^2(x^2+4x+13)) = 3/x + 1/x^2 + (2x + 5)/(x^2+4x+13))`

Integrating both sides yields:

`int (5x^3+18x^2+43x+13)/(x^2(x^2+4x+13)) dx= int 3/x dx+ ` `int 1/x^2 dx+ int (2x + 5)/(x^2+4x+13) dx`

You may use substitution to solve `int (2x + 5)/(x^2+4x+13) dx` such that:

`x^2+4x+13 = t => (2x + 4)dx = dt`

`int (2x + 5)/(x^2+4x+13) dx = int (2x + 4 + 1)/(x^2+4x+13)dx`

`int (2x + 5)/(x^2+4x+13) dx = int (2x + 4)/(x^2+4x+13) dx + int 1/(x^2+4x+13) dx`

`int (2x + 5)/(x^2+4x+13) dx = int dt/t +int 1/(x^2+4x+13) dx`

You should complete the square `x^2 + 4x` to solve the integral `int 1/(x^2+4x+13) dx` such that:

`x^2 + 4x + 4 - 4 = (x + 2)^2 - 4`

`x^2+4x+13 = (x + 2)^2 - 4 + 13`

`x^2+4x+13 = (x + 2)^2 + 9`

`int 1/(x^2+4x+13) dx = int 1/((x + 2)^2 + 9) dx`

`x + 2 = u => dx = du`

`int 1/((x + 2)^2 + 9) dx = int du/(u^2 + 9)`

`int du/(u^2 + 9) = (1/3) arctan (u/3) + c`

Substituting back `x + 2` for `u` yields:

`int 1/((x + 2)^2 + 9) dx = (1/3) arctan ((x+2)/3) + c`

`int (2x + 5)/(x^2+4x+13) dx = ln(x^2+4x+13) + (1/3) arctan ((x+2)/3) + c`

`int (5x^3+18x^2+43x+13)/(x^2(x^2+4x+13)) dx = 3ln|x| - 1/x + ln(x^2+4x+13) + (1/3) arctan ((x+2)/3) + c`

**Hence, evaluating the given integral yields** `int (5x^3+18x^2+43x+13)/(x^2(x^2+4x+13)) dx = 3ln|x| - 1/x + ln(x^2+4x+13) + (1/3) arctan ((x+2)/3) + c.`