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