Answer
$$\frac{1}{3}\ln \left| {\frac{{\sqrt {x + 9} - 3}}{{\sqrt {x + 9} + 3}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\sqrt {x + 9} }}} dx \cr
& {\text{Let }}x + 9 = {u^2},\,\,\,\,dx = 2udu \cr
& \int {\frac{1}{{x\sqrt {x + 9} }}} dx = \int {\frac{{2u}}{{\left( {{u^2} - 9} \right)u}}} du \cr
& = \int {\frac{2}{{{u^2} - 9}}} du \cr
& {\text{Decompose }}\frac{2}{{{u^2} - 9}}{\text{ into partial fractions}} \cr
& \frac{2}{{{u^2} - 9}} = \frac{2}{{\left( {u + 3} \right)\left( {u - 3} \right)}} \cr
& \frac{2}{{\left( {u + 3} \right)\left( {u - 3} \right)}} = \frac{A}{{u + 3}} + \frac{B}{{u - 3}} \cr
& {\text{Multiply by }}\left( {u + 3} \right)\left( {u - 3} \right){\text{ and simplify}} \cr
& 2 = A\left( {u - 3} \right) + B\left( {u + 3} \right) \cr
& {\text{if }}u = - 3,\,\,\,A = - 1/3 \cr
& {\text{if }}u = 3,\,\,\,B = 1/3 \cr
& \cr
& {\text{Replace the coefficients}} \cr
& \frac{2}{{{u^2} - 9}} = \frac{{ - 1/3}}{{u + 3}} + \frac{{1/3}}{{u - 3}} \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{2}{{{u^2} - 9}}} du = \int {\left( {\frac{{ - 1/3}}{{u + 3}} + \frac{{1/3}}{{u - 3}}} \right)} du \cr
& {\text{Integrate}} \cr
& = - \frac{1}{3}\ln \left| {u + 3} \right| + \frac{1}{3}\ln \left| {u - 3} \right| + C \cr
& = \frac{1}{3}\ln \left| {\frac{{u - 3}}{{u + 3}}} \right| + C \cr
& \cr
& {\text{Replace }}u = \sqrt {x + 9} \cr
& = \frac{1}{3}\ln \left| {\frac{{\sqrt {x + 9} - 3}}{{\sqrt {x + 9} + 3}}} \right| + C \cr} $$
