Answer
$$2{\tan ^{ - 1}}2x + \frac{{4x}}{{4{x^2} + 1}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{8dx}}{{{{\left( {4{x^2} + 1} \right)}^2}}}} \cr
& = 8\int {\frac{1}{{{{\left( {{{\left( {2x} \right)}^2} + 1} \right)}^2}}}dx} \cr
& {\text{Use trigonometric substitutions}} \cr
& {\text{ 2}}x = \tan \theta,{\text{ }}dx = \frac{1}{2}{\sec ^2}\theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& 8\int {\frac{1}{{{{\left( {{{\left( {2x} \right)}^2} + 1} \right)}^2}}}dx} = 8\int {\frac{1}{{{{\left( {{{\tan }^2}\theta + 1} \right)}^2}}}} \left( {\frac{1}{2}{{\sec }^2}\theta } \right)d\theta \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta = 1 + {\tan ^2}\theta \cr
& = 8\int {\frac{1}{{{{\left( {{{\sec }^2}\theta } \right)}^2}}}} \left( {\frac{1}{2}{{\sec }^2}\theta } \right)d\theta \cr
& {\text{simplifying, we get:}} \cr
& = 4\int {\frac{1}{{{{\sec }^2}\theta }}} d\theta = 4\int {{{\cos }^2}\theta } d\theta \cr
& = 4\int {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)} d\theta \cr
& = 2\int {\left( {1 + \cos 2\theta } \right)} d\theta \cr
& {\text{integrate}} \cr
& = 2\left( {\theta + \frac{1}{2}\sin 2\theta } \right) + C \cr
& = 2\left( {\theta + \sin \theta \cos \theta } \right) + C \cr
& = 2\theta + 2\sin \theta \cos \theta + C \cr
& {\text{write in terms of }}x,\cr
& \sin \theta = \frac{{2x}}{{\sqrt {4{x^2} + 1} }},\,\,\,\cr
& \cos \theta = \frac{1}{{\sqrt {4{x^2} + 1} }}{\text{ and }}\cr
& \theta = {\tan ^{ - 1}}2x \cr
& = 2{\tan ^{ - 1}}2x + 2\left( {\frac{{2x}}{{\sqrt {4{x^2} + 1} }}} \right)\left( {\frac{1}{{\sqrt {4{x^2} + 1} }}} \right) + C \cr
& = 2{\tan ^{ - 1}}2x + \frac{{4x}}{{4{x^2} + 1}} + C \cr} $$
