望采纳。
②换元x=tanu，dx=sec²udu
=∫√(ln(tanu+secu)+5)*secudu
=∫(ln(tanu+secu)+5)^(1/2)d(ln(tanu+secu)+5)
=(2/3)(ln(tanu+secu)+5)^(3/2)+C
③=∫√x(x+1)(√(x+1)-√x)dx
=∫(x+1)√x-(x+1-1)√(x+1)dx
=∫x^(3/2)+x^(1/2)-(x+1)^(3/2)+(x+1)^(1/2)dx
=(2/5)(x^(5/2)-(x+1)^(5/2))+(2/3)(x^(3/2)+(x+1)^(3/2))+C
④=∫u/sin³udcosu
=-∫ucsc²udu
=∫udcotu
=ucotu-∫cotudu
=ucotu-lnsinu+C
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