Question

求定积分 区间-2到2 ∫[(x+|x|)/(2+x^2)]dx,请给出详细过程,谢谢

Answer 1

当-2＜x＜0时
∫[(x+|x|)/(2+x^2)]dx=∫[(x-x)/(2+x^2)]dx=0

当0≤x＜2时

∫[(x+|x|)/(2+x^2)]dx=∫[(x+x)/(2+x^2)]dx=∫[2x/(2+x^2)]dx=∫[1/(2+x^2)]d(2+x^2)=ln(2+x^2)=ln6-ln2=ln3

所以在-2到2区间上
∫[(x+|x|)/(2+x^2)]dx=0+ln3=ln3

Answer 2

x小于零时函数等于0，所以变为0到2的积分，绝对值可去，函数变成2x/(2+x^2)，原函数是ln(2+x^2)，积分可求，答案是ln3

Answer 3

∫[-2,2][(x+|x|)/(2+x^2)]dx

=∫[-2,2]|x|/(2+x^2)dx
=2∫[0,2] x/(2+x^2)dx
=∫[0,2] 1/(2+x^2)dx^2
=ln(2+x^2) [0,2]
=ln6-ln2
=ln3

Answer 4

∫(-2 2) [(x+|x|)/(2+x^2)]dx
=∫(-2 0)[(x-x)/(2+x^2)]dx +∫(0 2)[2x/(2+x^2)]dx
=∫(-2 0) 0 dx +∫(0 2)[1/(x²+2)]d(x²)
=ln(x²+2)|(0 2)
=ln6 -ln2
=ln(6/2)
=ln3