Question

求不定积分√(x^2+a^2)/x^2dx

Answer 1

x=a*tant,t=arctan(x/a),dx=a*(sect)^2 *dt
原积分=Sa*sect/(a *tant)^2 *a*(sect)^2 *dt
=S(sect)^3 /(tant)^2 *dt
=S1/(cost*(sint)^2) dt
=Scost/((sint)^2 *(1-(sint)^2)dt
=S1/((sint)^2*(1-(sint)^2))dsint
(y=sint)
=S1/(y^2*(1-y^2))dy
=1/2*S(1/(1-y)+1/(1+y)dy+S1/y^2 *dy
=1/2*ln(1+y)/(1-y)-1/y+c
=1/2*ln(1+sint)/(1-sint)-1/sint+c
=1/2*ln(1+x/根号（x^2+a^2))/(1-x/根号（x^2+a^2))-根号（x^2+a^2) /x+c
=1/2*ln(根号（x^2+a^2)+x)/(根号（x^2+a^2)-x)-根号（x^2+a^2) /x+c