$ x^2 + y^2 − 6x − 4y + 5 = 0$

$(x-3)^2+(y-2)^2+5-9-4=0$

$(x-3)^2+(y-2)^2=(\sqrt(8))^2$

It is a circle with center at $(3,2)$ and radius = $\sqrt8$.

The points of intersection of circle with x axis are $(1,0)$ and $(5,0)$.

Rewrite the equation in terms of $\displaystyle\int y dx \\\displaystyle\sqrt{\left(8-(x-3)^2\right)}+2 = \int \sqrt{\left(8-(x-3)^2\right)} dx + \int2dx$ and so on..