$\int_0^3 \frac{\mathrm{~d} x}{(x+2) \sqrt{x+1}}=$

Evaluate the integral $\int_0^3 \frac{\mathrm{d}x}{(x+2)\sqrt{x+1}}$ using substitution.

Let $u = \sqrt{x+1}$, so $u^2 = x + 1$ and $x = u^2 - 1$. Differentiating gives $\mathrm{d}x = 2u\,\mathrm{d}u$. The limits transform from $x=0$ to $u=1$ and from $x=3$ to $u=2$.

Substituting into the integral:

$\int_1^2 \frac{2u\,\mathrm{d}u}{((u^2-1)+2)u} = \int_1^2 \frac{2u\,\mathrm{d}u}{(u^2+1)u} = \int_1^2 \frac{2\,\mathrm{d}u}{u^2+1}$

The antiderivative of $\frac{1}{u^2+1}$ is $\tan^{-1}(u)$. Therefore:

$2\left[\tan^{-1}(u)\right]_1^2 = 2\left(\tan^{-1}(2) - \tan^{-1}(1)\right)$

Using the identity $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ with $A=2$, $B=1$:

$\tan^{-1}(2) - \tan^{-1}(1) = \tan^{-1}\left(\frac{1}{3}\right)$

Thus, the integral simplifies to $2\tan^{-1}\left(\frac{1}{3}\right)$, which corresponds to option B.