求解三道微积分,积分符号不写了,1:根号下(e^x+1)dx 2:1/(x^2*根号下(1+x^2))dx.3:x^2 / 根号下(25-4x^2) dx
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 19:42:14
求解三道微积分,积分符号不写了,1:根号下(e^x+1)dx 2:1/(x^2*根号下(1+x^2))dx.3:x^2 / 根号下(25-4x^2) dx
求解三道微积分,积分符号不写了,
1:根号下(e^x+1)dx
2:1/(x^2*根号下(1+x^2))dx.
3:x^2 / 根号下(25-4x^2) dx
求解三道微积分,积分符号不写了,1:根号下(e^x+1)dx 2:1/(x^2*根号下(1+x^2))dx.3:x^2 / 根号下(25-4x^2) dx
2.
1/(x²√(1+x²))dx
=(1+x²-x²)/(x²√(1+x²))dx
=√﹙1+x²﹚/x²dx-1/√﹙1+x²﹚dx
=.
3
x²/√﹙25-4x²﹚dx
=﹙x²-25/4+25/4﹚/﹙25-4x²﹚
=.
相信接下的你会吧,欢迎追问
解法如下:
1 、设√e^x+1=t, 则e^x=t^2-1, x=ln(t^2-1) dx=2tdt/(t^2-1)∫√e^x+1 dx
=2∫ t^2/(t^2-1)dt
=2∫1+1/(t^2-1) dt
=2t+∫[1/(t-1 )-1/(t+1)]dt
=2t+ln(t-1)-ln(1+t)+c
根据t=(√e^x+1), 上...
全部展开
解法如下:
1 、设√e^x+1=t, 则e^x=t^2-1, x=ln(t^2-1) dx=2tdt/(t^2-1)∫√e^x+1 dx
=2∫ t^2/(t^2-1)dt
=2∫1+1/(t^2-1) dt
=2t+∫[1/(t-1 )-1/(t+1)]dt
=2t+ln(t-1)-ln(1+t)+c
根据t=(√e^x+1), 上式=2(√e^x+1)+ln[(√e^x+1)-1]-ln[(√e^x+1)+1]+c
2、∫1/(x^2*√(1+x^2) dx.
设x=tant , t∈(-π/2,π/2) 则 dx=(sect)^2 dt
∫1/(x^2*√(1+x^2) dx
=∫(sect)^2dt / [(tant)^2* sect]
= ∫sect dt /(tant)^2
=∫ csct*cott dt= - csct+c
根据tant=x, 则:csct= (√x^2+1)/x 所以上式= -(√x^2+1)/x +c
3、∫x^2 / √(25-4x^2) dx
设x=(5/2)sint t∈(-π/2,π/2) 则dx=(5/2)cost dt
∫x^2 / √(25-4x^2) dx
=∫[(25/4)(sint)^2/ 5cost ]*(5/2)cost dt
=(25/8)∫ (sint)^2 dt
=(25/16)∫(1-cos2t) dt
=(25/16)[t-(1/2)sin2t]+c
t=arcsin(2x/5) 另外由sint=2x/5, 可得 sin2t=2sint cost=(4x/5)*[√1-(4x^2/25)]
所以原式=(25/16)arcsin(2x/5)-(5x/8)√1-(4x^2/25) +c
!
收起
结果由mathematica 8生成
收起