Question

求根号下(x^2-9)/x的不定积分积分

Answer 1

令t=√(x^2-9),t^2=x^2-9,

2tdt=2xdx tdt=xdx

积分号下：√(x^2-9)dx/x

=√(x^2-9) xdx/x^2 （分子分母同乘以x）

=t *tdt/(t^2+9)

=t^2dt/(t^2+9)

=[1-9/(t^2+9)]dt

∫[1-9/(t^2+9)]dt=t-3arctan(t/3)+C=√(x^2-9)-3arctan[√(x^2-9)/3]+C