This integral can be done by using the tig substitution of x = sec(u)

Let's first change the operator and reevaluate the limits,

`x = sec(u)`

`dx = d(sec(u)) = sec(u) tan(u) du`

The limts are from x = 1 to x = 2,

Then the new limits are,

Bottom limit, x = 1

`1 = sec(u)` This gives,

`cos(u) = 1`

therefore when x = 1, u = 0

Upper limit, x = 2

`2 = sec(u)`

`cos(u) = 1/2`

Therefore when x = 2, u = `pi/3`

Now we will come to our integral,
 I will name it as 'I'

`I = int_1^2((5*sqrt(x^2-1))/(x))dx`

You can take the constant out of an integral, so,

`I = 5int_1^2((sqrt(x^2-1))/(x))dx`

Now we will make the substitution and change the operator and limits,

`I = 5int_0^(pi/3)((sqrt(sec^2(u)-1))/(sec(u)))sec(u) tan(u) du`

but we know the trignometric identity, `1 + tan^2(u) = sec^2(u)`

Therefore,

`I = 5int_0^(pi/3)((sqrt(tan^2(u)))/(sec(u)))sec(u) tan(u) du`

Simplifying,

`I = 5int_0^(pi/3)tan^2(u) du`

now again, using the same above identity,

`I = 5int_0^(pi/3)(sec^2(u)-1) du`

Separating the parts of the integral.

`I = 5(int_0^(pi/3)sec^2(u) du - int_0^(pi/3)du)`

Again changing the operator,

we know, `d(tan(u)) = sec^2(u) du`

Using that property, we can change the integral of the first part,

`I = 5(int_0^(pi/3)d(tan(u)) - int_0^(pi/3)du)`

`I = 5(tan(u) - u)` from u =0 to u = `pi/3` (I can't type the limits here)

this gives,

`I = 5((tan(pi/3) - tan(0)) - (pi/3 - 0) )`

`I = 5(tan(pi/3) - pi/3)`