Question

求不定积分∫(x^2+1)/(x^2—1)(x+1)dx

就是算这个分数的不定积分

Answer 1

答案在图片里

Answer 2

=∫(x^2-x+x+1)/(x^2-1)(x-1)dx
=∫x(x-1)/(x^2-1)(x-1)dx+∫(x+1)/(x^2-1)(x-1)dx
=∫x/(x^2-1)dx+∫1/(x-1)^2dx
=1/2(x^2-1)-1/(x-1)

追问

你题目看错了哦，分号下面是(x^2-1)(x+1)

追答

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