7-6x-x2 can be written as 7-(x2+6x+9-9).
Therefore,
7-(x2+6x+9-9)
= 16-(x2+6x+9)
= 16-(x+3)2
= (4)2-(x+3)2
∴ \(∫\)\(\frac {1}{\sqrt {7-6x-x^2}}\ dx\) = \(∫\frac {1}{\sqrt {(4)^2-(x+3)^2}} dx\)
Let x+3 = t
⇒ dx = dt
⇒ \(∫\frac {1}{\sqrt {(4)^2-(x+3)^2}} dx\) = \(∫\frac {1}{\sqrt {(4)^2-t^2}} dt\)
= \(sin^{-1}(\frac t4)+C\)
=\(sin^{-1}(\frac {x+3}4)+C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]

There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
These are tabulated below along with the meaning of each part.
