integration of ---1)∫(sin-1√x - cos-1√x)dx/ (sin-1 √x + cos-1√x)2) ∫√[(a-x)/(x-b)]dx3) ∫[√(x2 + 1)[log(x2 + 1) — 2log x] /x4]dx4) ∫(log (log x) + 1/(log x)2 )dx =x[f(x)-g(x)] +c --------f(x)=? and g(x)=?5) ∫[xlog(x+ √(1 + x2)) / √(1+x2) ] dx = A√(1 + x2)log(x + √(1 + x2)) + Bx + C-------A=? , B=?6) ∫cos{2tan-1√[(1-x) /(1+x)]}dx7) ∫[( ln x-1 ) / (1 + (ln x)2 )]dx8) ∫[(x4 + 1)/( x6 + 1 )] dx9) ∫ [x2 /(x4+1)]d[(x-1)/x]

∫[√(x² + 1)[log(x² + 1) — 2log x] /x⁴]dx

∫(x² + 1/x² )^1/2 [ log(x² + 1) - logx²/x³ ] dx

∫(1+ 1/x² )^1/2 [ log(1 + 1/x² )/x³ ]dx

let 1 + 1/x² = t²

      -2*1/x³ dx =2t dt

   1/x³ dx= -dt

                                     -∫ t logt² dt

                                    -∫ t (2logt) dt.  using by parts solve!!      

     not forgot 2   uproved

   ∫cos{2tan^-1√[(1-x) /(1+x)]}dx

     put  x = cos 2θ. 1-x = 2sin² θ and 1+x = 2cos² θ.

   u get inside by solving     2tan^-1 (tanθ ) = 2θ

       dx = -2sin2θ

     - ∫ 2cos 2θ *sin2θ dθ

     -  ∫ sin4θ dθ

      cos4θ/4 + c

     cos ² 2θ -1 /2 + c =  ANS    x²  -1 /2 +c