Step-by-step Solution:

1. Identify a suitable substitution. The integral is $I = \int \frac{\sec ^2 x}{(\sec x+\tan x)^{9 / 2}} d x$. The term $(\sec x+\tan x)$ in the denominator suggests the substitution: Let $t = \sec x + \tan x$.

2. Calculate the differential $dt$. Differentiating $t$ with respect to $x$, we get: $\frac{dt}{dx} = \frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x$ $\frac{dt}{dx} = \sec x (\tan x + \sec x) = \sec x \cdot t$ This gives us $dt = t \sec x \, dx$, which can be rearranged to $dx = \frac{dt}{t \sec x}$.

3. Express $\sec x$ in terms of $t$. We use the trigonometric identity $\sec^2 x - \tan^2 x = 1$. Factoring this difference of squares gives: $(\sec x - \tan x)(\sec x + \tan x) = 1$ Substituting $t = \sec x + \tan x$, we get: $(\sec x - \tan x) \cdot t = 1 \implies \sec x - \tan x = \frac{1}{t}$ Now we have a system of two linear equations: (i) $\sec x + \tan x = t$ (ii) $\sec x - \tan x = \frac{1}{t}$ Adding equations (i) and (ii): $2\sec x = t + \frac{1}{t} \implies \sec x = \frac{1}{2} \left(t + \frac{1}{t}\right)$

4. Substitute into the integral. We substitute $t$, $dx$, and $\sec x$ into the original integral: $I = \int \frac{\sec^2 x}{t^{9/2}} \cdot \frac{dt}{t \sec x} = \int \frac{\sec x}{t^{11/2}} dt$ Now, substitute the expression for $\sec x$ in terms of $t$: $I = \int \frac{\frac{1}{2} \left(t + \frac{1}{t}\right)}{t^{11/2}} dt = \frac{1}{2} \int \frac{t + t^{-1}}{t^{11/2}} dt$

5. Simplify and integrate. Simplify the integrand: $I = \frac{1}{2} \int (t \cdot t^{-11/2} + t^{-1} \cdot t^{-11/2}) dt = \frac{1}{2} \int (t^{-9/2} + t^{-13/2}) dt$ Now, integrate using the power rule, $\int u^n du = \frac{u^{n+1}}{n+1} + C$: $I = \frac{1}{2} \left[ \frac{t^{-9/2 + 1}}{-9/2 + 1} + \frac{t^{-13/2 + 1}}{-13/2 + 1} \right] + K$ $I = \frac{1}{2} \left[ \frac{t^{-7/2}}{-7/2} + \frac{t^{-11/2}}{-11/2} \right] + K$ $I = \frac{1}{2} \left[ -\frac{2}{7}t^{-7/2} - \frac{2}{11}t^{-11/2} \right] + K$ $I = -\frac{1}{7}t^{-7/2} - \frac{1}{11}t^{-11/2} + K$

6. Format the result to match the options. To match the form of the given options, we factor out the term with the lower power, which is $t^{-11/2}$. $I = -t^{-11/2} \left( \frac{1}{7} t^{-7/2 - (-11/2)} + \frac{1}{11} \right) + K$ $I = -t^{-11/2} \left( \frac{1}{7} t^{4/2} + \frac{1}{11} \right) + K$ $I = -t^{-11/2} \left( \frac{t^2}{7} + \frac{1}{11} \right) + K$

7. Substitute back for $t$. Replace $t$ with $\sec x + \tan x$: $I = -(\sec x + \tan x)^{-11/2} \left\{ \frac{1}{11} + \frac{1}{7}(\sec x + \tan x)^2 \right\} + K$ This can be written as: $I = -\frac{1}{(\sec x + \tan x)^{11/2}} \left\{ \frac{1}{11} + \frac{1}{7}(\sec x + \tan x)^2 \right\} + K$

8. Compare with the options. The derived expression matches option C.

   A: $-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^2\right\}+K$ (Incorrect sign inside the bracket) B: $\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^2\right\}+K$ (Incorrect signs) C: $-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+K$ (Correct) D: $\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+K$ (Incorrect overall sign)