重庆哪里按摩的比较多

重庆哪里按摩的比较多

1/(x^2 1)(x^2 x)=[(ax b)/(x^2 1)] (c/x) [d/(x 1)]

右边通分对应项相等，即可得到：

a=b=d=-1/2,c=1.

此时积分为：

原式

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

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

=-(1/4)∫d(x^2 1)/(x^2 1)-(1/2)arctanx-lnx-(1/2)ln(x 1)

=-(1/4)ln(1 x^2)-lnx-(1/2)ln(x 1)-(1/2)arctanx c

=-ln[(1 x^2)^(1/4)*x*(x 1)^(1/2)]-(1/2)arctanx c.