Question

高等数学求积分

Answer 1

Answer 2

=∫(x-1+6)/(x²-2x-1)dx
=1/2∫1/(x²-2x-1)d(x²-2x-1)+(3/√2)∫1/(x-1-√2)-1/(x-1+√2)dx
=(1/2)ln|x²-2x-1|+(3/√2)(ln|x-1-√2|-ln|x-1+√2|)+C

追答

三角换元脱根号，
1+x-x²=5/4-(x-1/2)²，令x=1/2+(√5/2)sinu

换元，令x=√2tanu

拆项
=∫1/x²+1/x+1/(1-x)dx
=ln|x|-ln|1-x|-1/x+C

三角换元脱根号，令x=(√6/3)tanu

三角变换，
=∫1/(2+2cos²(x/2-π/4)-1)dx
=∫sec²(x/2-π/4)/(sec²(x/2-π/4)+2)dx
=2∫1/(tan²(x/2-π/4)+3)dtan(x/2-π/4)
=(2/√3)arctan(tan(x/2-π/4)/√3)+C

追问

可以写个步骤嘛