put t =e^x/(x²+1)^1/2

     dt = 1/(x²+1)*(e^x.(x²+1)^1/2 - e^x.x/(x²+1)^1/2)

         = e^x(x²-x+1)/(x²+1)

I = integral of ( t.dt)

  =t²/2 + C

  = e^2x/2.(x²+1)   + C

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Ramesh
IIT Kgp - 05 batch

integral of e^x.(x^2-x+1)/(x^2+1)3/2

above expression can be rewritten as e^x.([x^2+1]/[x^2+1]^3/2 + [-x/(x^2+1)3/2])

integral of e^x(fx + dfx/dx) =e^x.fx +c

  so ans is e^x.1/(x^2+1)^1/2