If I=∫1+x21−x2dx then I is equal to

I=∫1+x21−x2  1+x21+x2dx=∫1+x2dx1−x21+x2I=∫1dx1−x21+x2+∫x2dx1−x21+x2I=∫dx1−x21+x2−∫−x2dx1−x21+x2I=∫dx1−x21+x2−∫1−x2−1dx1−x21+x2I=∫11−x21+x2dx−∫1−x2dx1−x21+x2+∫1dx1−x21+x2I=2∫dx1−x21+x2−∫dx1+x2I=2∫dxx21x2−1x1+1x2−log⁡x+1+x2Put 1x2+1=t2 in 1st  integral −2x3dx=dt2tdxx3=−tdt=I=2∫−tdtt2−2t−log⁡x+x2+1=I=2∫tdt(2)2−t2−log⁡x+x2+1+C=I=2⋅122log⁡2+t2−t−log⁡x+1+x2+C=I=12log⁡2+1x2+12−1x2+1−log⁡x+1+x2+C=I=12 log2x+1+x22x−1+x2 −logx+1+x2+C