求∫√(x-1)/xdx
2个回答

令√(x-1)=u,则x=u^2+1,dx=2udu

原式=∫ u/(u^2+1)*2udu

=2∫ u^2/(u^2+1)du

=2∫ (u^2+1-1)/(u^2+1)du

=2∫ 1du-2∫ 1/(u^2+1)du

=2u-2arctanu+C

=2√(x-1)-2arctan√(x-1)+C

数学软件验证,结果正确.

> int((x-1)^(1/2)/x,'x');

2*(x-1)^(1/2)-2*arctan((x-1)^(1/2))