Question

求不定积分过程加答案

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Answer 1

方法如下图所示，请认真查看，祝学习愉快：

Answer 2

图一(1)(3)(4)是积分基本公式
(2)∫1/x^2dx
=∫x^(-2)dx
=1/[1+(-2)]*x^(-2+1)+C
=-1/x+C(C为常数)
图二(1)∫cos2xdx
=(1/2)∫cos2xd(2x)
=(1/2)sin2x+C(C为常数)
(2)∫x/(1+x^2)dx
=(1/2)∫2x/(1+x^2)dx
=(1/2)∫1/(1+x^2)d(1+x^2)
=(1/2)In(1+x^2)+C(C为常数)
(3)∫cos√x/√xdx
=2∫cos√xd√x
=2sin√x+C(C为常数)
(4)∫dx/(x^2-x-2)=∫dx/(x-2)(x+1)
=(1/3)[∫dx/(x-2)-∫dx/(x+1)]
=(1/3)[In|x-2|-In|x+1|]+C=(1/3)In|(x-2)/(x+1)|+C(C为常数)