用x＝atant来做这道题求不定积分∫x^2⼀根号下(x^2 +a^2) dx (a>0)

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设x＝atant，则√(x^2+a^2)=asect,dx=a(sect)^2dt,
所以原式=∫（atant）^2*sectdt
=a^2∫(sint)^2dt/(cost)^3,
设u=sint,则du=costdt,
上式=a^2∫u^2du/(1-u^2)^2,
=(a^2/4)∫[u/(u-1)^2-u/(1+u)^2]du
=(a^2/4)∫[1/(u-1)+1/(u-1)^2-1/(u+1)+1/(u+1)^2]du
=(a^2/4)[ln|(u-1)/(u+1)}-1/(u-1)-1/(u+1)]+c
=(a^2/4)[ln|(u-1)/(u+1)|-2u/(u^2-1)]+c,
u=(x/a)/√[1+(x/a)^2]=x/√(x^2+a^2),
(u-1)/(u+1)=[x-√(x^2+a^2)/[x+√(x^2+a^2)]
=-[x-√(x^2+a^2)]^2/a^2,
2u/(u^2-1)=[2x/√(x^2+a^2)]/[x^2/(x^2+a^2)-1]
=-2x√(x^2+a^2)/a^2,
所以原式=(a^2/2){ln[√(x^2+a^2)-x]-lna}+x√(x^2+a^2)/2+c.