∫ 1/(x^2+9) dx
=(1/9)*∫ 1/((x/3)^2+1) dx
=(3/9)*∫ 1/((x/3)^2+1) d(x/3)
=(1/3)*arctan(x/3)+C
有不懂欢迎追问
你将1/(x^2+9)提一个1/9,(1/9)*(1/(x/3)^2+1),又arttanx=1/(x^2+1)
所以原式=arttan(x/3)/9+C(C为常数)
∫1/(x^2+9)dx
=1/9∫1/(x²/9+1)dx
=1/27∫1/[(x/3)²+1[ d(x/3)
=1/27*arctan(x/3)+C
令x=3t
dx=3dt
∫1/(x^2+9)dx
=∫3dt/(9t^2+9)
=(1/3)∫dt/(t^2+1)
=(1/3)arctan t+C
=(1/3)arctan (x/3)+C