Here are the solutions to the integrals from your quiz. I'll explain each step to ensure clarity.

1. $\int \left( z^2 + 2z + 1 \right)^{4} \, dz$
   Hint: $u = z + 1$

   Substitute $u = z + 1$, hence $dz = du$ and $z^2 + 2z + 1 = u^2$: $\int (z^2 + 2z + 1)^4 \, dz = \int u^8 \, du = \frac{u^9}{9} + C = \frac{(z+1)^9}{9} + C$

2. $\int \frac{ds}{2 - 7s}$

   Rewrite the integral using substitution $u = 2 - 7s$, hence $du = -7 \, ds$: $\int \frac{ds}{2 - 7s} = -\frac{1}{7} \int \frac{du}{u} = -\frac{1}{7} \ln|u| + C = -\frac{1}{7} \ln|2 - 7s| + C$

3. $\int x \cdot 10^{x^2 + 4} \, dx$

   Use substitution $u = x^2 + 4$, hence $du = 2x \, dx$: $\int x \cdot 10^{x^2+4} \, dx = \frac{1}{2} \int 10^u \, du = \frac{1}{2} \cdot \frac{10^u}{\ln(10)} + C = \frac{10^{x^2+4}}{2\ln(10)} + C$

4. $\int \frac{\sqrt{x} \, dx}{1 + x\sqrt{x}}$
   Hint: $u = 1 + x\sqrt{x}$

   Substituting $u = 1 + x\sqrt{x}$, so $du = \frac{3}{2}\sqrt{x} \, dx$: $\int \frac{\sqrt{x} \, dx}{1 + x\sqrt{x}} = \frac{2}{3} \int \frac{1}{u} \, du = \frac{2}{3} \ln|u| + C = \frac{2}{3} \ln|1 + x\sqrt{x}| + C$

5. $\int \sec \sec\left(\frac{2}{3}x\right) dx$

   The integral seems unclear due to the formatting, but clarify if it's involving sec functions or constants for better calculation.

6. $\int \sec^2(\arctan(\arctan x)) \frac{dx}{1+x^2}$
   Hint: $u = \arctan(\arctan x)$

   Substituting $u = \arctan(\arctan x)$, $du = \frac{dx}{1+x^2}$: [ \int \sec^2(\arctan(\arctan x)) \frac{dx}{1+x^2} = \int \sec^2(u) , du = \tan(u) + C =