∫x2−1x2+11+x4dx is equal to

I=∫x2−1x2+1x4+1dx=∫x21−1/x2x2(x+1/x)x2+1/x2dx=∫1−1/x2dx(x+1/x)(x+1/x)2−2 Putting x+1/x=t, we have I=∫dttt2−2 Again putting t2−2=y2,2tdt=2ydy, I=∫ydyy2+2y=12tan−1⁡y2=12tan−1⁡x2+1/x22+c