Evaluate ∫tan⁡θdθ = 12tan−1⁡tan⁡θ−12tan⁡θ+122log⁡AB+C

Let I=∫tan⁡θdθ  Let tan⁡θ=x2 . Then, d(tan⁡θ)=dx2 or sec2⁡θdθ=2xdx  or dθ=2x⋅dxsec2⁡θ=2xdx1+tan2⁡θ=2xdx1+x4I=∫x2×2xdx1+x4=∫2x2x4+1dx=∫2x2+1/x2dx=∫1+1/x2+1−1/x2x2+1/x2dx=∫1+1/x2x2+1/x2dx+∫1−1/x2x2+1/x2dx=∫1+1/x2(x−1/x)2+2dx+∫1−1/x2(x+1/x)2−2dxputting x−1x=u in the first integral and  x+1x=v in  the second integral we getI=∫duu2+(2)2+∫dvv2−(2)2=12tan−1⁡u2+122log⁡v−2v+2+C=12tan−1⁡x−1/x2+122log⁡x+1/x−2x+1/x+2+C=12tan−1⁡x2−1x2+122log⁡x2−x2+1x2+x2+1+C=12tan−1⁡tan⁡θ−12tan⁡θ=12tan−1⁡tan⁡θ−12tan⁡θ+122log⁡tan⁡θ−2tan⁡θ+1tan⁡θ+2tan⁡θ+1+C