Substitute [tex]u=x^2+4[/tex] and [tex]du=2x\,dx[/tex]. Then the integral transforms to
[tex]\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \int \frac{du}{u^3}[/tex]
Apply the power rule.
[tex]\displaystyle \int \frac{du}{u^3} = -\dfrac1{2u^2} + C[/tex]
Now put the result back in terms of [tex]x[/tex].
[tex]\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \left(-\dfrac1{2u^2} + C\right) = -\dfrac1{4u^2} + C = \boxed{-\dfrac1{4(x^2+4)^2} + C}[/tex]
