Question

求不定积分∫x^2/√(1+x^2) dx x的平方除以根号下1+x的平方 ∫x^2/√(1-x^2) dx x的平方除以根号下1-x的平

RT 求速解 谢谢 马上要考试了

Answer 1

见附图。

Answer 2

设x=tant,dx=(sec)^2dt,√[1+(tan)^2]=sect,
原式=∫（tanx)^2*(sect)^2dt/√[1+(tanx)^2]
=∫[(sec)t^2-1]sectdt
=∫(sect)^3dt-∫sectdt
∫(sect)^3dt=∫sectdtant
=sect*tant-∫tantd(sect)
=sect*tant-∫(tant)^2sectdt
=sect*tant-∫[(sect)^2-1]dt
=sect*tant-∫[(sect)^3dt+∫sectdt,
2∫(sect)^3dt=sect*tant+∫sectdt
∫(sect)^3dt=sect*tant/2+(1/2)ln|sect+tant|+C1,
原式=sect*tant/2+(1/2)ln|sect+tant|-ln|sect+tant+C
=(x/2)√（1+x^2)-(1/2)ln|x+√（1+x^2)|+C.

设x=sint,dx=costdt,√（1-x^2)=cost
原式=∫（sint)^2*costdt/cost
=(1/2)∫(1-cos2t)dt
=t/2-(1/4)∫cos2td(2t)
=t/2-(1/4)sin2t+C
=(1/2)t-(1/2)sintcost+C
=(1/2)arcsinx-(1/2)x√（1-x^2)+C.

Answer 3

见图

Answer 4

1/2*x*(1+x^2)^(1/2)-1/2*ln(x+(x^2+1)^(1/2))+c