Mark M. answered • 04/13/20

Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.

Recall: ∫[du / √[a^{2} - u^{2}] = Arcsin(u/a) + C

The given integral is equivalent to ∫dx / √[1 - (-7 - 6x +9x^{2})]

= ∫dx / √[1 - (9x^{2} - 6x + 1 - 8)] = ∫dx / √[9 - (3x-1)^{2}]

Let u = 3x-1, then du = 3dx. So, dx = (1/3)du

The integral then can be expressed as (1/3)∫[du / √[3^{2} - u^{2}]

= (1/3)Arcsin(u/3) + C = (1/3)Arcsin[(3x-1)/3] + C

Mark M.

04/13/20

Loisa M.

That's my answer too, (3x-1)^2 - 1, but how did it become, 9 - (3x-1)^2 (reffered to your first answer above)?04/14/20

Mark M.

04/15/20

Loisa M.

So, if the answer is (3x - 1)^2 - 1, then why is it your a=3, it should be a=1, right? So, therefore, the answer should be 1/3 arcsin 3x-1 + c only, it should not have 3 as denominator.04/15/20

Mark M.

04/15/20

Loisa M.

Is there really have 1/a beside of arcsin because in my book, there's none. And also i don't know how did you complete the square the quadratic, can you show me how?04/13/20