Isolate one of the square root terms.

Subtract $\sqrt{x-5}$ from both sides:

$$\sqrt{x+1} = 8 - \sqrt{x-5}$$

$$(\sqrt{x+1})^2 = (8 - \sqrt{x-5})^2$$ $$x+1 = 8^2 - 2(8)\sqrt{x-5} + (\sqrt{x-5})^2$$ $$x+1 = 64 - 16\sqrt{x-5} + (x-5)$$

Subtract $x$ from both sides:

$$1 = 59 - 16\sqrt{x-5}$$

Subtract 59 from both sides:

$$1 - 59 = -16\sqrt{x-5}$$ $$-58 = -16\sqrt{x-5}$$

Divide by -16:

$$\frac{-58}{-16} = \sqrt{x-5}$$ $$\frac{29}{8} = \sqrt{x-5}$$

$$\left(\frac{29}{8}\right)^2 = (\sqrt{x-5})^2$$ $$\frac{29^2}{8^2} = x-5$$ $$\frac{841}{64} = x-5$$

Add 5 to both sides:

$$x = \frac{841}{64} + 5$$

To add, find a common denominator:

$$x = \frac{841}{64} + \frac{5 \times 64}{64}$$ $$x = \frac{841}{64} + \frac{320}{64}$$ $$x = \frac{841 + 320}{64}$$ $$x = \frac{1161}{64}$$

Substitute $x = \frac{1161}{64}$ into the original equation:

$$\sqrt{\frac{1161}{64}+1} + \sqrt{\frac{1161}{64}-5} = 8$$ $$\sqrt{\frac{1161+64}{64}} + \sqrt{\frac{1161-320}{64}} = 8$$ $$\sqrt{\frac{1225}{64}} + \sqrt{\frac{841}{64}} = 8$$ $$\frac{\sqrt{1225}}{\sqrt{64}} + \frac{\sqrt{841}}{\sqrt{64}} = 8$$ $$\frac{35}{8} + \frac{29}{8} = 8$$ $$\frac{35+29}{8} = 8$$ $$\frac{64}{8} = 8$$ $$8 = 8$$

The solution is correct.

The value of $x$ is $\boxed{\frac{1161}{64}}$.