Integration by Parts.
It's not immediately obvious but this numerator differentiates very nicely.
[tex]\rm \left(x e^{2x}\right)'&=e^{2x}+2x e^{2x}=e^{2x}(2x+1)[/tex]
Notice that taking the derivative of the numerator actually creates a factor of our denominator.
Ah ha! So we've found a good choice for our u,
[tex]\rm u=x e^{2x}\qquad\qquad\qquad dv=\frac{1}{(2x+1)^2}~dx\\
\\
du=e^{2x}(2x+1)dx\qquad v=\frac{-1}{2(2x+1)}[/tex]
letting dv be everything else.
Applying Integration by parts
[tex]\rm =(u)(v)-\int (v)du[/tex]
gives us,
[tex]\rm =\left(x e^{2x}\right)\left(\dfrac{-1}{2(2x+1)}\right)-\int \left(\dfrac{-1}{2(2x+1)}\right)e^{2x}}(2x+1)dx[/tex]
Simplifying things a little bit before integrating again,
[tex]\rm =-\dfrac{x e^{2x}}{2(2x+1)}+\frac12\int e^{2x}~dx[/tex]
and integrating the last term,
[tex]\rm =-\dfrac{x e^{2x}}{2(2x+1)}+\frac14 e^{2x}[/tex]
looking for a common denominator, multiplying the first term by 2/2 and the second term by (2x+1)/(2x+1),
[tex]\rm =\dfrac{-2x e^{2x}}{4(2x+1)}+\dfrac{e^{2x}(2x+1)}{4(2x+1)}[/tex]
Combine the fractions together, factor out the exponential,
[tex]\rm =\dfrac{e^{2x}(-2x+2x+1)}{4(2x+1)}[/tex]
combine like-terms as a final step,
and include a constant of integration,
[tex]\rm =\dfrac{e^{2x}}{4(2x+1)}+c[/tex]
