1708\hspace{-16}$Let $\bf{I = \int\frac{\sec^2 x}{\left(\sec x +\tan x\right)^{\frac{9}{2}}}dx}$\\\\\\ Now Let $\bf{\left(\sec x+\tan x \right)=t.......................................................(1)}$\\\\\\ Then $\bf{\sec x \cdot \left(\sec x+\tan x \right)dx = dt}$\\\\\\ So $\bf{\left(\sec x \cdot t dx = dt\Rightarrow dx = \frac{1}{\sec x\cdot t}dt\right)}$\\\\\\ So $\bf{I = \int \sec ^2 x\cdot \frac{1}{\sec x}\cdot \frac{1}{t}\cdot \frac{1}{t^{\frac{9}{2}}}dt = \int \sec x \cdot \frac{1}{t^{\frac{11}{2}}}dt}$\\\\\\ Now Using $\bf{\bullet \left(\sec^2 x-\tan^2 x \right) = 1\Rightarrow \left(\sec x-\tan x\right)=\frac{1}{t}...........(2)}$\\\\\\ So from equation.......$\bf{(1)}$ and $\bf{(2)\;,}$ We get\\\\\\ $\bf{\Rightarrow 2\sec x = t+\frac{1}{t}=\frac{t^2+1}{2}\Rightarrow \sec x = \frac{t^2+1}{2t}}$\\\\\\ So $\bf{I = \frac{1}{2}\int\frac{t^2+1}{t^{\frac{11}{2}}}dt = \frac{1}{2}\int t^{-\frac{7}{2}}+\frac{1}{2}\int t^{-\frac{11}{2}}dt}$\\\\\\
\hspace{-16}$So $\bf{I = \frac{1}{2}\cdot -\frac{7}{2}t^{\frac{-5}{2}}+\frac{1}{2}\cdot -\frac{11}{2}t^{\frac{-9}{2}}+\mathbb{C}}$\\\\\\ So $\bf{I = \int\frac{\sec^2 x}{\left(\sec x +\tan x\right)^{\frac{9}{2}}}dx=\frac{-1}{4}\left\{7\left(\sec x+\tan x\right)^{-\frac{5}{2}}}+11\left(\sec x+\tan x\right)^{-\frac{-9}{2}} \right\}+\mathbb{C}$