Help me with this beta gamma integral

[CristianTheodor]

I don't know what substitution to make , maybe you can help . (I must solve using beta gamma integral functions)
[MATH]\int_{0}^{\infty }\frac{x}{(4x^2-2x+1)^2}=? \\ \int_{0}^{\infty }\frac{x}{(4x^2-2x+1)^2}=\int_{0}^{\infty }\frac{16x}{((2x-1)^2+3)^2}[/MATH]

 

[CristianTheodor]

making this substitution doesn't work :
[MATH][/MATH] \[u=(2x-1)\] \[x=(u+1)/2\]
\[dx=\frac{1}{2}du\] and I get this \[\int_{-1}^{\infty}\frac{4(u+1)du}{(u^2+3)^2}\] which I cannot solve