(3/2)∫ 1/[(x-1/2)²+3/4] dx=?

∫(1-x)^2/x^3 dx ∫(x-1)^2/x^3 dx ∫x^3/1+x^2 dx ∫(X^3)/(1+X^2)dx ∫(3x^4+x^2)/(x^2+1)dx x-9/[(根号)x]+3 dx ∫ x+1/[(根号)x] dx ∫ [（3-x^2）]^2 dx 计算 ∫（x^4-2x^3+x^2+1）/x（x-1)² dx ∫(x-4x^2+4x^3) (1-x^2)^(1/2) dx=-4∫x^2 * (1-x^2)^(1/2)dx,其中x 不定积分 ∫1/（x²-3x+2）dx 求∫ dx/ (x²-x+1)^(3/2) ∫1/[(2-3x)(2x+1)] dx= ∫(1-x^2)^(3/2)dx/(x^4)