Assuming you mean \(\sec2x+\tan2x\), you have
\(\displaystyle\int(\sec2x+\tan2x)\,\mathrm dx=\int\sec2x\frac{\sec2x+\tan2x}{\sec2x+\tan2x}\,\mathrm dx+\int\frac{\sin2x}{\cos2x}\,\mathrm dx\)
\(\displaystyle=\int\frac{\sec^22x+\sec2x\tan2x}{\sec2x+\tan2x}\,\mathrm dx+\int\frac{\sin2x}{\cos2x}\,\mathrm dx\)
For the first integral, set \(u=\sec2x+\tan2x\), so that \(\mathrm du=2(\sec2x\tan2x+\sec^22x)\,\mathrm dx\); for the second, \(v=\cos2x\), so that \(\mathrm dv=-2\sin2x\,\mathrm dx\). Then you have
\(\displaystyle\frac12\int\frac{\mathrm du}u-\frac12\int\frac{\mathrm dv}v\)
\(\displaystyle=\frac12(\ln|u|-\ln|v|)+C\)
\(\displaystyle=\ln\left|\frac uv\right|^{1/2}+C\)
\(\displaystyle=\ln\left|\frac{\sec2x+\tan2x}{\cos2x}\right|^{1/2}+C\)
\(\displaystyle=\ln\left|\sec^22x(1+\sin2x)\right|^{1/2}+C\)
