$\displaystyle \int\ (dx/4+x^{2})^{2}$
dont know how to do this problem without the square root
$\displaystyle \int{\frac{dx}{(4 + x^2)^2}}$
$\displaystyle x = 2\tan{t}$
$\displaystyle dx = 2\sec^2{t} \, dt$
$\displaystyle \int \frac{2\sec^2{t}}{(4 + 4\tan^2{t})^2} \, dt$
$\displaystyle \frac{1}{8} \int \frac{\sec^2{t}}{(1 + \tan^2{t})^2} \, dt$
$\displaystyle \frac{1}{8} \int \frac{\sec^2{t}}{(\sec^2{t})^2} \, dt$
$\displaystyle \frac{1}{8} \int \frac{1}{\sec^2{t})} \, dt$
$\displaystyle \frac{1}{8} \int \cos^2{t} \, dt$
take it from here?