用换元积分法求解！
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解：令x^(1/4)=y，则x=y^4，dx=4y^3dy
原式=∫y^2/(y^2-y)*4y^3dy
=∫4(y^4-1+1)/(y-1)dy
=∫4(y^4-1+1)/(y-1)dy
=4∫[y³+y²+y+1+1/(y-1)]dy
=4[y^4/4+y³/3+y²/2+y+ln|y-1|]+C
=x+4/3*x^(3/4)+2x^(1/2)+4x^(1/4)+4ln|x^(1/4)-1|+C
原式=∫y^2/(y^2-y)*4y^3dy
=∫4(y^4-1+1)/(y-1)dy
=∫4(y^4-1+1)/(y-1)dy
=4∫[y³+y²+y+1+1/(y-1)]dy
=4[y^4/4+y³/3+y²/2+y+ln|y-1|]+C
=x+4/3*x^(3/4)+2x^(1/2)+4x^(1/4)+4ln|x^(1/4)-1|+C
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