Let’s solve the two integrals step by step:

(u) $\int \frac{dx}{\sqrt{8x - x^2}}$:

1. Rewriting the quadratic expression:
   Factor $8x - x^2$ as $-\left(x^2 - 8x\right)$. Completing the square for $x^2 - 8x$, we get:
   $x^2 - 8x = (x - 4)^2 - 16.$
   So: $8x - x^2 = 16 - (x - 4)^2.$

   Substituting this, the integral becomes: $\int \frac{dx}{\sqrt{16 - (x - 4)^2}}.$

2. Recognizing the form:
   This is of the form $\int \frac{dx}{\sqrt{a^2 - (x - c)^2}}$, which evaluates to:
   $\arcsin\left(\frac{x - c}{a}\right) + C.$

3. Solution:
   Here, $a = 4$ and $c = 4$. Therefore: $\int \frac{dx}{\sqrt{8x - x^2}} = \arcsin\left(\frac{x - 4}{4}\right) + C.$

(v) $\int \frac{e^t}{\sqrt{e^{2t} + 9}} \, dt$:

1. Substitution:
   Let $u = e^t$, so that $du = e^t dt$. The integral becomes: $\int \frac{du}{\sqrt{u^2 + 9}}.$

2. Recognizing the form:
   This is of the form $\int \frac{dx}{\sqrt{x^2 + a^2}}$, which evaluates to: $\ln\left|x + \sqrt{x^2 + a^2}\right| + C.$

3. Solution:
   Here, $x = u$ and $a = 3$. Substituting back $u = e^t$, we get: $\int \frac{e^t}{\sqrt{e^{2t} + 9}} \, dt = \ln\left|e^t + \sqrt{e^{2t} + 9}\right| + C.$

Would you like more details or any clarifications?

5 Related Questions:

1. How do you complete the square for other quadratic expressions in integrals?
2. What are other examples of arc-trigonometric substitutions in integrals?
3. Can you derive the general form of $\int \frac{dx}{\sqrt{a^2 - x^2}}$?
4. How do logarithmic solutions appear in integrals involving $\sqrt{x^2 + a^2}$?
5. How can you verify your solution using differentiation?

Tip:

Always check if completing the square simplifies the integral before attempting substitutions or transformations!