∫√（2x+1）dx

∫x√（1+x）dx∫x√（1+x）dx∫x√（1+x）dx∫x√（1+x）dx=∫(1+x)^(3/2)dx-∫√(1+x)dx=(2/5)(1+x)^(5/2)-(2/3)x^(3/2)+C答案2

∫x√(1+2x)dx∫x√(1+2x)dx∫x√(1+2x)dx这个是考你的换元能力来的,~~~~不明白的就追问吧~~~~希望楼主采纳!O(∩_∩)O谢谢高数～我不会

求几个微积分解答∫（2x+1）³dx,∫（x+1）／√xdx,∫㏑²x／xdx求几个微积分解答∫（2x+1）³dx,∫（x+1）／√xdx,∫㏑²x／xdx求几

x-9/[(根号)x]+3dx∫x+1/[(根号)x]dx∫[（3-x^2）]^2dxx-9/[(根号)x]+3dx∫x+1/[(根号)x]dx∫[（3-x^2）]^2dxx-9/[(根号)x]+3d

∫1/√（-x^2+2x）dx∫1/√（-x^2+2x）dx∫1/√（-x^2+2x）dx

求不定积分∫1/（x^2√x）dx求不定积分∫1/（x^2√x）dx求不定积分∫1/（x^2√x）dx令t=√x,则x=t²,dx=2tdt∴∫1/(x²√x)dx=∫(1/t^5

∫lnx/x√（x^2-1）dx∫lnx/x√（x^2-1）dx∫lnx/x√（x^2-1）dx解答见下面图片：

求不定积分（1）∫dx/sin^2（2）∫(1+√x)^2dx求不定积分（1）∫dx/sin^2（2）∫(1+√x)^2dx求不定积分（1）∫dx/sin^2（2）∫(1+√x)^2dx∫dx/sin

∫1/[x(1+√x)^2dx∫1/[x(1+√x)^2]dx∫1/[x(1+√x)^2dx∫1/[x(1+√x)^2]dx∫1/[x(1+√x)^2dx∫1/[x(1+√x)^2]dx原式=∫2√x

∫x√x+1dx(x根号x+1dx）求不定积分.∫x√x+1dx(x根号x+1dx）求不定积分.∫x√x+1dx(x根号x+1dx）求不定积分.令√(x+1)=u,则x=u²-1,dx=2u

∫dx/(1+√(1-x^2))∫dx/(1+√(1-x^2))∫dx/(1+√(1-x^2))答：设x=sint原式=∫[1/(1+cost)]d(sint)=∫[cost/(1+cost)]dt=

∫dx/[√(2x-1)+1]∫dx/[√(2x-1)+1]∫dx/[√(2x-1)+1]令a=√(2x-1)+1x=(a²-2a+2)/2所以dx=(a-1)da所以原式=∫(a-1)da

∫lnx/√（x+1）dx∫lnx/√（x+1）dx∫lnx/√（x+1）dxxxInot

∫（√1+e^x）dx∫（√1+e^x）dx∫（√1+e^x）dx令√(1+e^x)=u,则e^x=u^2-1,x=ln(u^2-1),dx=2udu/(u^2-1)I=∫√(1+e^x)dx=∫2u

∫dx／（√x）-1∫dx／（√x）-1∫dx／（√x）-1令√x=t那么得到原积分=∫1/(t-1)d(t^2)=∫2t/(t-1)dt=∫2+2/(t-1)dt=2t+2ln|t-1|+C=2√x

∫(x+1)*√(2-x2)dx∫(x+1)*√(2-x2)dx∫(x+1)*√(2-x2)dx如图

∫dx/√(x+1)^2+9.∫dx/√(x+1)^2+9.∫dx/√(x+1)^2+9.令x+1=3tanθ,则dx=3sec²θdθ∫1/√[(x+1)²+9]dx=∫1/√(

∫dx/√[1-e^(-2x)]∫dx/√[1-e^(-2x)]∫dx/√[1-e^(-2x)]∫dx/√[1-e^(-2x)]lete^(-x)=siny-e^(-x)dx=cosydy∫dx/√[

∫dx/(√(x+1)+2∫dx/(√(x+1)+2∫dx/(√(x+1)+2令√x+1=t,那么x=t^2-1所以原积分=∫1/(t+2)d(t^2-1)=∫2t/(t+2)dt=∫2-4/(t+2

∫√1-x^2dx∫√1-x^2dx∫√1-x^2dx令x=sint则dx=costdt原式=∫costcostdt=0.5∫(1+cos2t)dt=0.5[t+0.5sin2t]+C=0.5[arc