Evaluate $$I=\int\frac{1}{x^2-8x+8}~dx$$
First, complete the square using the denominator: $$x^2-8x+8=(x-4)^2-8$$ and therefore, let $x=2\sqrt{2}\sec{\theta}+4$, $\therefore dx=2\sqrt{2}\sec{\theta}\tan{\theta}~d\theta$, and hence, we have $$I=2\sqrt{2}\int\frac{\sec{\theta}\tan{\theta}}{(2\sqrt{2}\sec{\theta}+4-4)^2-8}~d\theta$$ $$=2\sqrt{2}\int\frac{\sec{\theta}\tan{\theta}}{(2\sqrt{2}\sec{\theta})^2-8}~d\theta$$ $$=2\sqrt{2}\int\frac{\sec{\theta}\tan{\theta}}{8\sec^2{\theta}-8}~d\theta$$ $$=\frac{\sqrt{2}}{4}\int\frac{\sec{\theta}\tan{\theta}}{\sec^2\theta-1}~d\theta$$ $$=\frac{\sqrt{2}}{4}\int\frac{\sec{\theta}\tan{\theta}}{\tan^2\theta}~d\theta$$ $$=\frac{\sqrt{2}}{4}\int\frac{\sec{\theta}}{\tan\theta}~d\theta$$ note:$$\frac{\sec{\theta}}{\tan{\theta}}=\frac{\cos{\theta}}{\sin{\theta}}\cdot\frac{1}{\cos{\theta}}=\csc{\theta}$$ therefore, $$I=-\frac{\sqrt{2}}{4}\ln{|\csc\theta+\cot\theta|}+C$$ now, using the fact that $x=2\sqrt{2}\sec\theta+4$, we have the following identities, based on the definition of the trigonometric functions: $$\csc\theta=\frac{x-4}{\sqrt{x^2-8x+8}}$$ as well as $$\cot\theta=\frac{2\sqrt{2}}{\sqrt{x^2-8x+8}}$$ therefore, the final integral is given by: $$I=-\frac{\sqrt{2}}{4}\ln{\left|\frac{x+2\sqrt{2}-4}{\sqrt{x^2-8x+8}}\right|}+C$$ which, I've been informed, is incorrect. Would anyone be kind enough to help me realize my error? Any responses are appreciated. Thank you.