The integral of 1/((h-r)^2+r^2) from 0 to h is arctan((h-r)/r) evaluated at h and 0, which simplifies to arctan(h/r) - arctan(-r/r) or arctan(h/r) + pi/2.
To solve this integral, you can use trigonometric substitution. Let r = h*tan(u) and solve for dr in terms of u. Then substitute these expressions into the integral and simplify to get the final answer.
The domain of the function 1/((h-r)^2+r^2) is all real numbers except for when the denominator is equal to 0. In other words, the domain is all real numbers except for r = h.
The range of the function 1/((h-r)^2+r^2) is all real numbers greater than or equal to 0. This is because the numerator is always 1, and the denominator is always positive, so the function can never be negative.
The graph of 1/((h-r)^2+r^2) from 0 to h is a smooth, continuous curve. It starts at 0 when r = 0, increases as r increases, and approaches infinity as r approaches h. There are no vertical asymptotes, and the curve is always above the x-axis.