√1-x²dx

∫1/[(√X)(1+X)]dx

∫dx/x+√(1-x²)

∫1/√x*(4-x)dxLog就是ln的意思.后面自己加一个常数C即可.

∫dx/[x√(1-x^4)]∫dx/[x√(1-x^4)]letx^2=siny2xdx=cosydy∫dx/[x√(1-x^4)]=(1/2)∫(1/siny)dy=(1/2)ln|cscy-coty|+C=(1/2)ln|1/x^2-

不定积分dx/√x+x^1/3

不定积分：∫√(x+1)/x)dx若是I=∫[√(x+1)/x]dx,令√(x+1)=t,则x=t^2-1,I=∫[√(x+1)/x]dx=∫2t^2dt/(t^2-1)=2∫[1+1/(t^2-1)]dt=2t+∫[1/(t-1)-1/(

求不定积分x√(1+x)dx令a=√(1+x)x=a²-1dx=2ada所以原式=∫(a²-1)*a*2ada=2∫(a^4-a²)da=2a^5/5-2a³/3+C=2(1+x)²√(1+

求不定积分√(x-1)/xdx∫[√(x-1)/x]dxletx=(secy)^2dx=2secytanydy∫[√(x-1)/x]dx=∫2(tany)^2/(secy)dy=2∫(siny)^2/cosydy=2∫(1-(cosy)^2

积分号dx/(√x)(1+x)令t=√xx=t^2dx=2tdt∫dx/(√x)(1+x)=∫2tdt/t(1+t^2)=2∫dt/(1+t^2)=2arctant+C=2arctan(√x)+C积分dx/(√x)(1+x)=积分2d(√x

dx/((√x)(1+x))的不定积分

∫dx/x(1+√x)∫dx/x(1+√x)令1+√x=t,x=(t-1)^2dx=2(t-1)dt∫dx/x(1+√x)=∫2(t-1)dt/[t(t-1)^2]=2∫dt/[t(t-1)]会了吧

∫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2/3*x(x+1)^(3/2)-4/15*(x+1)^(5/2)+C过程我只写第

x√(1-x)dx怎求换元,令t=√(1-x),则x=1-t^2,x√(1-x)dx=(-2t^2+2t^4)dt求出t的代数式后,带入t=√(1-x)

∫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

∫1/[x(1+√x)^2dx∫1/[x(1+√x)^2]dx原式=∫2√xd(√x)/[(√x)²(1+√x)²]=2∫d(√x)/[√x(1+√x)²]=2∫[1/√x-1/(1+√x)-1/(1+√x)&

∫x√x+1dx(x根号x+1dx）求不定积分.令√(x+1)=u,则x=u²-1,dx=2udu原式=∫(u²-1)*u*2udu=2∫(u^4-u²)du=(2/5)u^5-(2/3)u³+C=(

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

∫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/a=∫(1-1/a(da=a-ln|a|+C'=√(2x-1)+1-ln[√(2x-1)+1]+C

∫arctan(1+√x)dx∫arctan(1+√x)dx换元t=arctan(1+√x),(tant-1)^2=x=∫td(tant-1)^2=t(tant-1)^2-∫(tant-1)^2dt=t(tant-1)^2-∫(sint-c

∫√(1-x²)dxx=sinadx=cosada√(1-x²)=cosa∴原式=∫cos²ada=1/2∫(cos2a+1)da=1/4·sin2a+1/2a+c=1/2x√(1-x²)+1/2ar