Question

求不定积分？

Answer 1

解：设√x=t，则x=t²，dx=2tdt 原式=2∫tsintdt =-2∫tdcost =-2tcost+2∫costdt =-2tcost+2sint+C =-2√xcos√x+2sin√x+C

Answer 2

I=∫[x^2/√(x^2+a^2)]dx =∫[(x^2+a^2-a^2)/√(x^2+a^2)]dx =∫√(x^2+a^2)dx-∫(a^2)/√(x^2+a^2)]dx =x√(x^2+a^2)-I-(a^2)ln[x+√(x^2+a^2)]+2C, 得 I=∫[x^2/√(x^2+a^2)]dx=(1/2)x√(x^2+a^2)-(a^2/2)ln[x+√(x^2+a^2)]+C。

Answer 3

令√x=t--->x=2t,dx=2tdt 原式=arctant*2tdt/[t(1+t^2)] =2arctantdt/(1+t^2) =(arctant)^2+C =(aectan√x)^2+C

Answer 4

x/(1+x)^3＝(1+x－1)/(1+x)^3＝1/(1+x)^2 － 1/(1+x)^3。 所以， ∫x/(1+x)^3dx＝∫[1/(1+x)^2 － 1/(1+x)^3]d(1+x)＝－1/(1+x)+1/2*(1+x)^2+C

Answer 5

∫[In(sinx)]dx/(sinx)^2 =∫[In(sinx)]d(-cotx) =-(cotx)In(sinx)+∫cotxd(In(sinx)) =-(cotx)In(sinx)+∫cotxcosx/sinxdx =-(cotx)In(sinx)+∫(cotx)^2dx =-(cotx)In(sinx)+∫[(cscx)^2-1]dx =-(cotx)In(sinx)-cotx-x+C。