Question

数学不定积分和极限题求解

Answer 1

=lim(3-x-1-x)/(x²-1)(√(3-x)+√(1+x))
=lim-2/(x+1)(√(3-x)+√(1+x))
=-2/2*(√2+√2)
=-√2/4

追答

=∫tan²usecudtanu
=∫(secu)^5-sec³udu
其中∫(secu)^5du=∫sec³udtanu
=sec³utanu-∫tanu(3sec²utanusecu)du
=sec³utanu-3∫tan²usec³udu
=sec³utanu+3∫sec³udu-3∫(secu)^5du
=(sec³utanu+3∫sec³udu)/4
所以定积分=(sec³utanu-∫sec³udu)/4
=(2sec³utanu-secutanu-ln|tanu+secu|)/8+C