∫1/(x^4-x^2)dx

来源：学生作业帮编辑：大师作文网作业帮分类：数学作业时间：2024/11/10 22:50:49

∫1/(x^4-x^2)dx

原式=∫[(1/2)/(x-1)-(1/2)/(x+1)-1/x²]dx
=(1/2)ln│x-1│-ln│x+1│+1/x+C (C是积分常数)
=(1/2)ln│(x-1)/(x+1)│+1/x+C.

∫（x^2+1/x^4)dx ∫1/(x^4-x^2)dx ∫（(x+2)/4x(x^2-1)）dx 积分∫x/[(x^2+1)(x^2+4)]dx ∫(3x^4+x^2)/(x^2+1)dx 求不定积分 ∫ 1/(1+2x)² dx ∫ x/√x²+4 dx ∫1/(x^2+x+1)dx ∫X^2/1-x^2 dx. ∫dx/x^2(1-x^2) ∫dx/x^2(1+x^2) ∫X^4/1+x dx. ∫dx/x(1+x^4)