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i am using $ sign for integration

Let p=(4x²-3x+2)^1/2  =[(2x-3/4)²+23/16]^1/2

=>$p=(2x-3/4)/2.( 4x²-3x+2)^1/2 +23/32log[(2x-3/4)+p]

1/8[(8x-3)p+23/4log[(2x-3/4)+p]=r(say)

 I=int[(x^3-7x^2-4x+2)/p)dx]+int[(8x-3)/pdx]

Let I1= int[(8x-3)/pdx]

Let p^2=t         => 8x-3dx=dt

=> I1=int[(1/t^1/2)dt]=2t^1/2=2p

=> I=int[(x^3-7x^2-4x+2)/p)dx]+2p

Let I2= int[(x^3-7x^2-4x+2)/p)dx]

=(1/4)[int[(4x^3-27x^2+20x-12)/p)dx]-int[(x^2-36x+20)/pdx]]

=(1/4)[int(x-6)pdx- int[(x^2-36x+20)/pdx]]              (dividing)

=(1/4)[int(xp)-6r- int[(x^2-36x+20)/pdx]]

int(xp)=(1/8)int(8xpdx)

=(1/8)[int[(8x-3)pdx]+3int(p)]

Solving similar to I1   int(xp)=(1/8)[(2/3)p^3+3r]

int[(x^2-36x+20)/pdx]]=(1/4) int[(4x^2-144x+80)/pdx]]

=(1/4)[int(p)-int[(141x-78)/pdx]         (dividing)

 =(1/4)[r-17I1-int[(5x-26)/pdx]           (dividing (141x-78)by p)

=(1/4)[r-34p-(5/8)[int[(8x-3)/pdx]-(193/8)int(1/p)]]

=(1/4)[r-34p-(5/8)[2p-(193/8r)]]

=>I2=(1/4)[(1/8)[(2/3)p^3+3r]-6r-(1/4)[ r-34p-(5/8)[2p-(193/8r)]]

=>I=(1/4)[(1/8)[(2/3)p^3+3r]-6r-(1/4)[ r-34p-(5/8)[2p-(193/8r)]]+2p

      =(1/32)[(2/3)p^3-29r-2[ r-34p-(5/8)[2p-(193/8r)]]+2p

      =(1/32)[(2/3)p^3-31r+68p+(5/4)[ 2p-(193/8r)]]+2p

Really long!!!