求1/(4+9x^2)的不定积分 求1/(4x^2+4x+5)的不定积分

Question

如题所述

Answer 1

详细解答如下，点击放大：

Answer 2

∫[1/(4+9x²)]dx:
令x=(2/3)tanθ,dx=(2/3)sec²θdθ
∫[1/(4+9x²)]dx
=∫{[(2/3)sec²θ]/4+9[(2/3)tanθ]²}dθ
=∫{[(2/3)sec²θ]/(4+4tan²θ)}dθ
=∫{[(2/3)sec²θ]/4sec²θ}dθ
=∫[(2/3)/4]dθ
=(1/6)θ+C ，∵x=(2/3)tanθ,θ=arctan(3x/2)
∴=(1/6)arctan(3x/2)