Evaluate the integral integral27 (1)/((x+1)sqrt(x+2)) dx.

Let $u = \sqrt{x+2}$.
Then $u^2 = x+2$, which means $x = u^2 - 2$.
From this, $x+1 = (u^2 - 2) + 1 = u^2 - 1$.
Differentiating $u^2 = x+2$ with respect to $x$, we get $2u\,du = dx$.

When $x=2$, $u = \sqrt{2+2} = \sqrt{4} = 2$.
When $x=7$, $u = \sqrt{7+2} = \sqrt{9} = 3$.

Let $\frac{2}{(u-1)(u+1)} = \frac{A}{u-1} + \frac{B}{u+1}$.
Multiplying by $(u-1)(u+1)$ gives:

Set $u=1$: $2 = A(1+1) + B(1-1) \implies 2 = 2A \implies A=1$.
Set $u=-1$: $2 = A(-1+1) + B(-1-1) \implies 2 = -2B \implies B=-1$.
So, the integrand becomes:

Using the logarithm property $\ln a - \ln b = \ln \frac{a}{b}$:

Again, using the logarithm property $\ln a - \ln b = \ln \frac{a}{b}$:

The final answer is $\ln\frac{3}{2}$.