求不定积分[(x^2+9)^2]分之一dx
问题描述：
求不定积分[(x^2+9)^2]分之一dx
答
∫dx/(x^2+9)
=(1/3)∫d(x/3)/[(x/3)^2+1]
=(1/3)arctan(x/3) +C题目是 ∫dx/[(x^2+9）^2]∫dx/[(x^2+9)^2]x=3tanutanu=x/3 cos2u=2(cosu)^2-1=2/(1+tanu^2)-1=(1-tanu^2)/(1+tanu^2) tan2u=2tanu/(1-tanu^2) sin2u=tan2u*cos2u=2tanu/(1+tanu^2)=2x/3/(1+x^2/9)=6x/(9+x^2)dx=3(secu)^2du(x^2+9)^2=81secu^2=∫(1/27)du/(secu)^2=∫(1/27)*(cosu)^2du=∫(1/54)(1+cos2u)du=(1/54)u+(1/108)sin2u+C=(1/54)arctan(x/3)+(1/18)(x)/(9+x^2) +C