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upasana ghosh Grade: 12
        

integration of ---1)          ∫(sin-1√x - cos-1√x)dx/ (sin-1 √x + cos-1√x)


                       2)           ∫√[(a-x)/(x-b)]dx


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


                       4)          ∫(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)]}dx


                       7)          ∫[( ln x-1 ) / (1 + (ln x)2 )]dx


                       8)          ∫[(x4 + 1)/( x6 + 1 )] dx


                       9)          ∫ [x2 /(x4+1)]d[(x-1)/x]

6 years ago

Answers : (2)

gOlU g3n|[0]uS
42 Points
										

∫[√(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 Smile  uproved

6 years ago
gOlU g3n|[0]uS
42 Points
										

   ∫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    x-1 /2 +c

6 years ago
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