Question

求下面这个不定积分，需要详细过程，谢谢～

Answer 1

设x=tanu
dx=sec²u du
1+x²=1+tan²u=sec²u
(1)当secu＞0时，√(1+x²)=secu
原式=∫sec²u du/secu
=∫secu du
=∫du/cosu
=∫cosudu/cos²u
=∫dsinu/cos²u
=∫dsinu/(1-sin²u)
=(1/2)[∫dsinu/(sinu+1)-∫dsinu/(sinu-1)]
=(1/2)(ln|sinu+1|-ln|sinu-1|)+C
=(1/2)ln|(sinu+1)/(sinxpu-1)|+C (对数里分子分母都乘以sinu+1)
=(1/2)ln|(sinu+1)²/cos²u|+C
=ln|(sinu+1)/cosu|+C
=ln|tanu+secu|+C
而tanu=x，secu=√(x²+1)
所以
原式=ln|x+√(x²+1)|+C
(2)当secu＜0时，√(1+x²)=-secu
原式=-ln|tanu+secu|+C
=-ln|x-√(1+x²)|+C
=ln[|x-√(1+x²)|^(-1)]+C
=ln|x+√(1+x²)|+C

Answer 2

简解如图: