∫tan−1⁡xdx is equal to

I=∫1⋅tan−1⁡xdx=tan−1⁡x⋅x−∫11+x⋅12x⋅xdx=xtan−1⁡x−12∫x1+xdxLet I1=∫xx+1dxAssume that x=t⇒ 12xdx=dt⇒dx=2tdt∴ I1=∫t⋅2t1+t2dt=2∫1+t21+t2−11+t2dt =2t−2tan−1⁡t =2x−2tan−1⁡x∴ I=xtan−1⁡x−122x−2tan−1⁡x+C =xtan−1⁡x−x+tan−1⁡x+C =(x+1)tan−1⁡x−x+C