∫1/[(2-3x)(2x+1)] dx=

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/11/20 16:11:47

∫1/[(2-3x)(2x+1)] dx=
答案是1/7ln|(2x+1)/2-3x|+c

∫1/[(2-3x)(2x+1)] dx
=∫6/[(4-6x)(6x+3)] dx
=∫6/7[1/(4-6x)+1/(6x+3)] dx
=1/7[∫1/(6x+3) d(6x+3)+∫1/(4-6x)d(4-6x)]
=1/7ln(6x+3)-ln(4-6x)+C
=1/7ln[(6x+3)/(4-6x)]+C

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