Step-by-Step Solution

1. Substitution: Let $u = \sqrt{x}$. Then, $du = \frac{1}{2\sqrt{x}} dx$, which implies $dx = 2\sqrt{x} \, du = 2u \, du$.
2. Rewrite the integral: Substituting into the integral, we get: $\int \frac{1}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}+2)} \, dx = \int \frac{1}{u(u+1)(u+2)} 2u \, du = 2 \int \frac{1}{(u+1)(u+2)} \, du$
3. Partial Fraction Decomposition: We can express $\frac{1}{(u+1)(u+2)}$ as $\frac{A}{u+1} + \frac{B}{u+2}$. Multiplying both sides by $(u+1)(u+2)$, we get $1 = A(u+2) + B(u+1)$.
   • Setting $u = -1$, we get $1 = A(-1+2) + B(0)$, so $A = 1$.
   • Setting $u = -2$, we get $1 = A(0) + B(-2+1)$, so $B = -1$.
   Thus, $\frac{1}{(u+1)(u+2)} = \frac{1}{u+1} - \frac{1}{u+2}$.
4. Integrate: Now we have: $2 \int \frac{1}{(u+1)(u+2)} \, du = 2 \int \left(\frac{1}{u+1} - \frac{1}{u+2}\right) \, du = 2 \left(\int \frac{1}{u+1} \, du - \int \frac{1}{u+2} \, du\right)$ $= 2 (\ln|u+1| - \ln|u+2|) + C = 2 \ln\left|\frac{u+1}{u+2}\right| + C$
5. Substitute back: Replace $u$ with $\sqrt{x}$: $2 \ln\left|\frac{\sqrt{x}+1}{\sqrt{x}+2}\right| + C$