Answer
$\frac{1}{2}\ln \left( {\frac{{18}}{{13}}} \right) - \frac{\pi }{6} + \frac{2}{3}\arctan \left( {\frac{2}{3}} \right)$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{x}{{{x^2} + 4x + 4 + 9}}} dx \cr
& \int_0^1 {\frac{x}{{{{\left( {x + 2} \right)}^2} + 9}}} dx \cr
& {\text{Integrate by substitution}} \cr
& {\text{Let }}u = x + 2,{\text{ }}du = dx \cr
& {\text{The new limitsof integration are}} \cr
& x = 1 \to u = 3 \cr
& x = 0 \to u = 2 \cr
& {\text{Substituting}} \cr
& \int_0^1 {\frac{x}{{{{\left( {x + 2} \right)}^2} + 9}}} dx = \int_2^3 {\frac{{u - 2}}{{{u^2} + 9}}du} \cr
& = \int_2^3 {\left( {\frac{u}{{{u^2} + 9}} - \frac{2}{{{u^2} + 9}}} \right)du} \cr
& {\text{Integrating}} \cr
& = \left[ {\frac{1}{2}\ln \left( {{u^2} + 9} \right) - \frac{2}{3}\arctan \left( {\frac{u}{3}} \right)} \right]_2^3 \cr
& {\text{Evaluating}} \cr
& {\text{ = }}\left[ {\frac{1}{2}\ln \left( {{{\left( 3 \right)}^2} + 9} \right) - \frac{2}{3}\arctan \left( {\frac{3}{3}} \right)} \right] \cr
& - \left[ {\frac{1}{2}\ln \left( {{{\left( 2 \right)}^2} + 9} \right) - \frac{2}{3}\arctan \left( {\frac{2}{3}} \right)} \right] \cr
& {\text{ = }}\frac{1}{2}\ln \left( {18} \right) - \frac{2}{3}\left( {\frac{\pi }{4}} \right) - \frac{1}{2}\ln \left( {13} \right) + \frac{2}{3}\arctan \left( {\frac{2}{3}} \right) \cr
& {\text{ = }}\frac{1}{2}\ln \left( {\frac{{18}}{{13}}} \right) - \frac{\pi }{6} + \frac{2}{3}\arctan \left( {\frac{2}{3}} \right) \cr
& \approx 0.031114 \cr} $$
