这道第二类换元法积分题该怎样解?
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这道第二类换元法积分题该怎样解?
![](http://img.wesiedu.com/upload/b/f2/bf222ecb7f7189c4b8687c52c27e32e1.jpg)
![](http://img.wesiedu.com/upload/b/f2/bf222ecb7f7189c4b8687c52c27e32e1.jpg)
![这道第二类换元法积分题该怎样解?](/uploads/image/z/1653200-8-0.jpg?t=%E8%BF%99%E9%81%93%E7%AC%AC%E4%BA%8C%E7%B1%BB%E6%8D%A2%E5%85%83%E6%B3%95%E7%A7%AF%E5%88%86%E9%A2%98%E8%AF%A5%E6%80%8E%E6%A0%B7%E8%A7%A3%3F)
∫dx/(x+根号1-x^2) ,不用换元法很难积分.
令x=sint
原式=∫cost/(sint+cost) dt
=1/2 ∫(cost-sint)/(sint+cost) dt+1/2 ∫(cost+sint)/(sint+cost) dt
=1/2∫1/(sint+cost) d(sint+cost)+1/2∫dt
=1/2ln|sint+cost|+1/2t+c
t=arcsinx
cost=√1-x^2
原式=1/2ln|x+√1-x^2|+1/2arcsinx+c
令x=sint
原式=∫cost/(sint+cost) dt
=1/2 ∫(cost-sint)/(sint+cost) dt+1/2 ∫(cost+sint)/(sint+cost) dt
=1/2∫1/(sint+cost) d(sint+cost)+1/2∫dt
=1/2ln|sint+cost|+1/2t+c
t=arcsinx
cost=√1-x^2
原式=1/2ln|x+√1-x^2|+1/2arcsinx+c