We are asked to evaluate two integrals: a) $\int \frac{1}{(x+3)(x+6)} dx$ b) $\int \frac{1}{(x^{1/2})(x^{1/2})} dx$

1. Problem Description

We are asked to evaluate two integrals:

a) $\int \frac{1}{(x+3)(x+6)} dx$

b) $\int \frac{1}{(x^{1/2})(x^{1/2})} dx$

2. Solution Steps

a) We can use partial fraction decomposition to evaluate the integral.

$\frac{1}{(x+3)(x+6)} = \frac{A}{x+3} + \frac{B}{x+6}$

$1 = A(x+6) + B(x+3)$

Let $x = -3$:

$1 = A(-3+6) + B(-3+3)$

$1 = 3A$

$A = \frac{1}{3}$

Let $x = -6$:

$1 = A(-6+6) + B(-6+3)$

$1 = -3B$

$B = -\frac{1}{3}$

Thus,

$\frac{1}{(x+3)(x+6)} = \frac{1/3}{x+3} - \frac{1/3}{x+6}$

Now we can integrate:

$\int \frac{1}{(x+3)(x+6)} dx = \int \left( \frac{1/3}{x+3} - \frac{1/3}{x+6} \right) dx$

$= \frac{1}{3} \int \frac{1}{x+3} dx - \frac{1}{3} \int \frac{1}{x+6} dx$

$= \frac{1}{3} \ln|x+3| - \frac{1}{3} \ln|x+6| + C$

$= \frac{1}{3} (\ln|x+3| - \ln|x+6|) + C$

$= \frac{1}{3} \ln \left| \frac{x+3}{x+6} \right| + C$

b) The integral is $\int \frac{1}{x^{1/2} x^{1/2}} dx = \int \frac{1}{x} dx$.

$\int \frac{1}{x} dx = \ln|x| + C$

3. Final Answer

a) $\frac{1}{3} \ln \left| \frac{x+3}{x+6} \right| + C$

b) $\ln|x| + C$