Question

integral: sqrt (16 + 6x − x^2)
dx

Answer 1

Note that since (x-3)^2 = x^2-6x+9, what you have is

∫√(25-(x-3)^2) dx
If u = x-3, du = dx, and it's just
∫√(25-u^2) du

which you should know. If not, just use the trig substitution u = 5sinθ

The answer can be seen at

http://www.wolframalpha.com/input/?i=%E2%88%AB%E2%88%9A%2825-u^2%29+du

Or, in terms of the original integral,

http://www.wolframalpha.com/input/?i=%E2%88%AB%E2%88%9A%2816%2B6x-x^2%29+dx

Answer 2

thanks. i wish i could upload a screen shot of what i typed it but i did it exactly as the site gave and it was counted wrong.

Answer 3

nevermind! i just forgot the constant