∫ 1/√(x²-2x+10) dx=∫ 1/√[(x-1)²+9] dx令(x-1)=3tanu,则√[(x-1)²+9]=3secu,dx=3sec²udu=∫ sec²u/secu du=∫ secu du=ln|secu+tanu| + C=ln|√[(x-1)²+9]/3+(x-1)/3| + C=ln[√...