Question

这两种不定积分怎样算？求详细过程。谢谢啦。

Answer 1

第一个 ，三角代换
令x=asect,dx=sect tant dt
原式=∫atant*d(asect)
=a^2∫tant sect tant dt (1)
=a^2∫tant d sec t
=a^2tant sect -a^2∫sect sec^2 t dt
=a^2tant sect -a^2∫ sect(tan^2 t+1) dt=a^2tant sect -a^2∫sect dt -a^2∫tant sect tant dt (2)
联立(1),(2)所以
a^2∫tant sect tant dt
=a^2/2[tant sect -∫sect dt]
=a^2/2[tant sect- ln|sect+tan t| ]+C
=a^2/2*x√(x²-a²)-a^2/2ln|x+√(x²-a²)|+c
第二个，分部积分
√(x^2+a^2)dx=
x√(x^2+a^2)-∫xd√(x^2+a^2)
=x√(x^2+a^2)-∫x^2/√(x^2+a^2)dx
=x√(x^2+a^2)-∫(x^2+a^2-a^2)/√(x^2+a^2)dx
=x√(x^2+a^2)-∫[√(x^2+a^2)-a^2/√(x^2+a^2)]dx
移项,得
2∫√(x^2+a^2)dx
=x√(x^2+a^2)+a^2∫1/√(x^2+a^2)dx
=x√(x^2+a^2)+a^2ln|x+√(x^2+a^2)|+2c
所以原式=
1/2 x√(x^2+a^2)+1/2 a^2ln|x+√(x^2+a^2)|+c
也可设x=atanu，则u=arctan(x/a)计算