Question

求不定积分∫1/(x^2+4)dx求高手解题要步骤谢谢

Answer 1

凑微分，和换元法人家两个人都说了
三角代换
设x = 2 tan u
dx = 2 sec^2 u du

∫1/(x^2+4)dx
= ∫2sec^2 u/(4sec^2 u) du
= (1/2)u + c

= (1/2) tan^-1(x/2) + c
凑微分直接公式
∫1/(x^2+4)dx

=1/4∫1/[1+(x/2)²]dx
=1/2∫1/[1+(x/2)²]d(x/2)
=1/2arctanx/2+c

Answer 2

∫1/(x^2+4)dx

=1/4∫1/[1+(x/2)²]dx
=1/2∫1/[1+(x/2)²]d(x/2)
=1/2arctanx/2+c

Answer 3

设x = 2 tan u
dx = 2 sec^2 u du

∫1/(x^2+4)dx
= ∫2sec^2 u/(4sec^2 u) du
= (1/2)u + c

= (1/2) tan^-1(x/2) + c

Answer 4

晕，反正切函数求导公式你会吧？这个书上有公式的