Evaluating indefinite integral: $ \int { \frac { 1 }{ \sqrt { 8-{ x }^{ 2 }-2x } } } \,dx$

Question

How can i evaluate this indefinite integral.

$$ \int { \frac { 1 }{ \sqrt { 8-{ x }^{ 2 }-2x } } } \,dx$$

I know it involves completing square but i don't know how to do it.

Answer 1

Hint:

$$8 - x^2 - 2x = 8 - (x^2 + 2x) = 8 - (x^2 + 2x + 1) + 1 = 9 - (x + 1)^2$$

Can you take it from here?

Answer 2

Hints:

$$8-x^2-2x=9-(x+1)^2\implies \sqrt{8-x^2-2x}=3\sqrt{1-\left(\frac{x+1}3\right)^2}$$

and now do remember the derivative of $\;\arcsin\;$ ...