integration 👍👍😉?

Luckily for you, you don't need to integrate.  You just have to find the function for the normal.  The normal to a curve at a point is perpendicular to the tangent to a curve at that point

y = (x - 1)^(-2) + 2

y' = -2 * (x - 1)^(-3)

x = 2

y' = -2 * (2 - 1)^(-3)

y' = -2 / 1^3

y' = -2

The tangent would have a slope of -2, so the normal would have a slope of -1/(-2), or 1/2

Now just use point-slope form to find the rest.  We know that the normal passes through (2 , 3)

y - 3 = (1/2) * (x - 2)

y - 3 = (1/2) * x - 1

y = (1/2) * x + 2

Hint: This has nothing to do with integration. Whats the slope of the curve at point A?

Thus, whats the slope of the normal there? Surely you can take it from there. 

y = 2 + (x-1)^(-2), y' = -2(x-1)^(-3). At A(2,3) y' = -2(2-1)^(-3) = -2. slope

of normal at A = -1/(-2) = (1/2). Equation of normal through A is given by

(y-3) = (1/2)(x-2), ie., 2y-6 = x-2, ie., y = (1/2)x +2