如题：∫(x-1)/(x^2+1)dx

来源：学生作业帮编辑：百度作业网作业帮分类：数学作业时间：2024/06/13 08:30:32

如题：∫(x-1)/(x^2+1)dx

∫(x-1)/(x^2+1)dx
=∫x/(x^2+1)dx-∫1/(x^2+1)dx
=1/2∫1/(x^2+1)d(x^2+1)-arctanx
=1/2ln(x^2+1)-arctanx+C

如题：∫(x-1)/(x^2+1)dx ∫dx/(x²+1)(x²+x) 如题,求不定积分 ∫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^2+1/x^4)dx ∫(1-x)^2/x^3 dx ∫ x/(1+X^2)dx= ∫2 -1|x²-x|dx ∫(x-1)^2/x^3 dx ∫x√(1+2x)dx