You should come up with the following substitution, such that:

`x^2 - 5 = t^2 => 2xdx = 2tdt => dx = (tdt)/(sqrt(t^2+5))`

`sqrt(x^2 - 5) = sqrt t^2 = t`

Changing the variable, yields:

`int 1/((t^2+5)*t))*(tdt)/(sqrt(t^2+5))`

You may use partial fraction decomposition, such that:

`1/(t(t^2 + 5)) = A/t + (Bt + C)/(t^2+5)`

`1 = At^2 + 5A + Bt^2 + Ct`

Equating the coefficients of like powers yields:

`{(A+B = 0),(C = 0),(5A = 1):} => A=-B=1/5`

`1/(t(t^2+5)) = 1/(5t) - 1/(5(t^2+5))`

Integrating both sides yields:

`int 1/((t^2+5)*t))*(tdt)/(sqrt(t^2+5)) = (1/5)int (dt)/(sqrt(t^2+5)) - (1/5)int tdt/((t^2+5)sqrt(t^2+5))`

You should come up with the substitution `t^2 + 5 = u` , such that:

`2t dt = du => tdt = (du)/2`

`int tdt/((t^2+5)sqrt(t^2+5)) = (1/2)int du/(usqrt u)`

`int tdt/((t^2+5)sqrt(t^2+5)) = (1/2)int (du)/(u*u^(1/2))`

`int tdt/((t^2+5)sqrt(t^2+5)) = (1/2)int (du)/(u^(1/2+1))`

`int tdt/((t^2+5)sqrt(t^2+5)) = (1/2)int u^(-3/2)du`

`int tdt/((t^2+5)sqrt(t^2+5)) = (1/2) u^(-3/2+1)/(-3/2+1)`

`int tdt/((t^2+5)sqrt(t^2+5)) = (1/2) u^(-1/2)/(-1/2) = -1/(sqrtu)`

Substituting back `t^2 + 5` for u yields:

`int tdt/((t^2+5)sqrt(t^2+5)) = -1/(sqrt(t^2+5))`

`int 1/((t^2+5)*t))*(tdt)/(sqrt(t^2+5)) = (1/5)int (dt)/(sqrt(t^2+5)) + 1/(sqrt(t^2+5))`

`int 1/((t^2+5)*t))*(tdt)/(sqrt (t^2+5)) = (1/5)ln(t + sqrt(t^2 + 5)) + 1/(sqrt(t^2+5)) + c`

**Hence, evaluating the given integral yields**` I =
(1/5)ln(sqrt(x^2 - 5) + x) + 1/x + c`

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