Step 1: Substitute u = e^x first:
Let $u = e^{x}$, so $du = e^{x}dx$. The limits become $u = 1$ to $u = 10$. The integrand $\frac{e^x\sqrt{e^x-1}}{e^x+8}dx$ turns into $\frac{\sqrt{u-1}}{u+8}\,du$.
Step 2: Second substitution:
Now let $w = \sqrt{u-1}$, so $u = w^2 + 1$ and $du = 2w\,dw$. The limits become $w = 0$ to $w = 3$. Also $u + 8 = w^2 + 9$. The integral is $\int_0^3 \frac{2w^2}{w^2+9}\,dw$.
Step 3: Evaluate with a standard result:
Use $\frac{w^2}{w^2+9} = 1 - \frac{9}{w^2+9}$. Then $\int_0^3 \frac{2w^2}{w^2+9}dw = 2\cdot 3 - 18\int_0^3\frac{dw}{w^2+9}$. We know $\int_0^3\frac{dw}{w^2+9} = \frac{1}{3}\tan^{-1}1 = \frac{\pi}{12}$. So the value is $6 - 18\cdot\frac{\pi}{12} = 6 - \frac{3\pi}{2}$.
Step 4: Sanity check against the options:
Numerically $6 - 4.712 = 1.288$. The integrand at $x = 1$ is about $0.33$, at $x = 2$ about $1.21$ and at $x = \log 10$ about $1.67$, so the area over a length of $2.3$ is of the order of $1.3$. That agrees with $\frac{3}{2}(4-\pi) = 1.29$. Option (A) is about $10.7$ and option (D) is about $5.4$, far too large. Option (C) is about $0.64$, too small.
Final Answer:
The integral equals $6 - \frac{3\pi}{2} = \frac{3}{2}(4-\pi)$, option (B).
\[ \boxed{\frac{3}{2}(4-\pi)} \]