Integration DX upon root 16 - 6 x-
x square
geez - use math for calculus problems, ok? You want
∫ dx/(16-6x-x^2) = ∫ dx/(7 - (x+3)^2)
This is almost a standard form, so let u = x+3 and you have
∫ du/(7-u^2)
Now you can use partial fractions to get
1/(2√7) ∫ 1/(√7+x) + 1/(√7-x) dx
and those are just logs when you integrate.
Or, you can express it as a hypebolic function, since
∫ du/(a^2-u^2) = -arctanh(u/a)