On this problem, we're going to make it into something that looks like Pythagorean identity. So we're going to complete the square for the Y ^2 -, 8 Y and taking half of the -8 and squaring it half a -8 negative 4 ^2 is 16. So this plus 32 is really going to get split up into plus 16 and another plus 16 so that these first three terms are going to factor in the Y, the quantity y -, 4 ^2 + 16. Now this is a Pythagorean identity. It's a ^2 + b ^2. So our opposite side is going to be y - 4 and our adjacent is going to be 4. So we get tangent Theta is y - 4 / 4. Multiply the four across and take the derivative of each side SO4 secant squared Theta D Theta equal DY. When we look at the integral for the DY, we're going to put in that four secant squared Theta D Theta. The denominator y -, 4 quantity squared is really just going to be this four tangent Theta squared or 16 tangent squared Theta plus 16. Factoring out the 16, we get tangent squared Theta plus one, which we know is secant squared Theta. So the secant squared thetas are both going to cancel leaving just the integral of 1/4 D Theta which is 1/4 Theta plus C. Going back to this original up here, this original substitution, we can see that Theta is going to be tangent inverse of y - 4 / 4 + C So our final answer 1/4 tangent inverse y - 4 / 4 plus the C.