∫dx /(9+25x^2)

解：∫dx /(9+25x^2)
=(1/9)*∫dx/[1+(5x/3)^2]
=(1/9)*(3/5)∫d(5x/3)/[1+(5x/3)^2]
=(1/15)*∫du/(1+u^2)
=（1/15）*arctan(u)+c
=(1/15)*arctan(5x/3)+c

注意：∫du/(1+u^2)
=∫d(tanA)/[1+(tanA)^2]
=∫[(secA)^2/(secA)^2]*dA
=A+c
=arctan(u)+c

∫dx 除以括号9加25x的平方