Question

求不定积分 ∫[(√x^2+a^2)/x^2]dx

Answer 1

解：原式=-∫√(x^2+a^2)d(1/x)
=-√(x^2+a^2)/x+∫(1/x)d[√(x^2+a^2)]
=-√(x^2+a^2)/x+∫(1/x)*(1/2)*2x/√(x^2+a^2)dx
=-√(x^2+a^2)/x+∫dx/√(x^2+a^2)
=-√(x^2+a^2)/x+ln|x+√(x^2+a^2)|+C，其中C是任意常数

Answer 2

∫[(√x^2+a^2)/x^2]dx=∫a^2(sect)^3/(tant)^2dx[tant=x/a,dx=ad(tant)=a(sect)^2dt]=∫a^2/(sint^2*cost)dt=a^2∫1/costd(cott)=a^2[sint-∫sint/(cost^2)*cottdt]=a^2[sint-∫1/costdt]=a^2(sint+ln(sect-tant))+c
t=arctan(x/a)