Question

求定积分∫(上限为2，下限为1) 根号(x^2-1) dx/x

Answer 1

先求不定积分∫√(x^2-1)/xdx=∫√(1-x^-2)dx; 设x^-2=u^2; dx=-udu/x^-3; ∫√(1-X^2)dx=-∫u√(1-u^2)du/(x^-3)=(1-u^2)^(3/2)/3x^3+C=(1-x^-2)^(3/2)/3x^3+C。再把积分区间代入就行了。