Answer
$$\frac{1}{2}\ln \left( {17} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^4 {\frac{x}{{{x^2} + 1}}dx} \cr
& {\text{substitute }}u = {x^2} + 1,{\text{ }}du = 2xdx \cr
& {\text{express the limits in terms of }}u \cr
& x = 0{\text{ implies }}u = {\left( 0 \right)^2} + 1 = 1 \cr
& x = 4{\text{ implies }}u = {\left( 4 \right)^2} + 1 = 17 \cr
& {\text{the entire integration is carried out as follows}} \cr
& \int_0^4 {\frac{x}{{{x^2} + 1}}dx} = \int_1^{16} {\frac{{1/2}}{u}du} \cr
& {\text{find the antiderivative}} \cr
& = \left. {\left( {\frac{1}{2}\ln \left| u \right|} \right)} \right|_1^{17} \cr
& {\text{Use the fundamental theorem}} \cr
& = \frac{1}{2}\ln \left| {17} \right| - \frac{1}{2}\ln \left| 1 \right| \cr
& {\text{Simplify}} \cr
& = \frac{1}{2}\ln \left( {17} \right) - \frac{1}{2}\ln \left( 1 \right) \cr
& = \frac{1}{2}\ln \left( {17} \right) \cr} $$
