求1 (x^3 4x)dx的不定积分
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∫dx/[x+x^(n+1)]=∫dx/[x(1+x^n)]=∫[(1+x^n)-x^n]dx/[x(1+x^n)]=∫dx/x-∫x^(n-1)dx/(1+x^n)=lnx-x^n/n!+C再问:最
令(1-x)/x=t^2,则:1-x=xt^2,∴(1+t^2)x=1,∴x=1/(1+t^2),∴dx=[2t/(1+t^2)^2]dt.∴∫{1/√[x(1-x)]}dx=∫{[(1-x)+x]/
我想LZ的意思是求不定积分:∫(e^x)/(1+e^2x)dx=∫1/(1+e^2x)d(e^x)然后用第二类换元法,令e^x=tant,则t=arctan(e^x)代入可得:∫1/(1+e^2x)d
请采纳再答:
∫[√(x-1)/x]dxletx=(secy)^2dx=2secytanydy∫[√(x-1)/x]dx=∫2(tany)^2/(secy)dy=2∫(siny)^2/cosydy=2∫(1-(co
令√x=tx=t方,dx=2tdt所以原式=∫2tdt/[(1+t方)t]=2∫1/(1+t方)dt=2arctant+c=2arctan√x+c
1∫(1/x)sin(lnx)dx=∫sin(lnx)dlnx=-cos(lnx)+C2∫3^(-x/2)dx=-2*3^(-x/2)/ln3+C3∫(x+1)f'(x)dx=f(x)*(x+1)-∫
∫(x+1)e∧xdx=∫(x+1)de∧x=(x+1)e∧x-∫e∧xd(x+1)=(x+1)e∧x-e∧x=xe∧x
利用倒代换即设x=1/t,dx=-1/t^2dt则原式为-(积分号)t/(t-1)dt即-(积分号)dt-(积分号)d(t-1)/(t-1)得-t-ln|t-1|+C再代换回来得-1/x-ln|1/x
答:设t=√x,x=t^2∫[1/(√x+x)]dx=∫[1/(t+t^2)]d(t^2)=∫[1/(t+t^2)]2tdt=2∫[1/(1+t)]d(t+1)=2ln(t+1)+C=2ln(√x+1
用第一换元法即可……设u=e^x,则du=e^xdx,积分即可化为∫du/(1-u^2)=-1/2ln|(u-1)/(u+1)|=-1/2ln|(e^x-1)/(e^x+1)|
sysxabf1=x+1;f2=0.5*x^2;int(f1,0,1)+int(f2,1,2)f=exp(ax)*sin(bx)inf(f)
看图片:\x0d\x0d
x⁴/(x²+1)=x²(x²+1-1)/(x²+1)=x²-x²/(x²+1)=x²-(x²+1
结果无法用初等函数表示,用浏览器算了一下,结果如下:
答案有错,∫x^2/(1-x^4)dx=∫x^2/(1-x^4)dx=∫x^2/[(1+x^2)(1-x^2)]dx=∫1/2[1/(1-x^2)-1/(1+x^2)]dx=1/2∫[1/(1-x^2
∫1/[x(x^2+1)]dx,d(x^2)=2xdx=(1/2)∫1/[x^2*(x^2+1)]d(x^2)=(1/2)∫1/x^2d(x^2)-(1/2)∫1/(x^2+1)d(x^2)=(1/2
令x=1/t先求∫lnt/(1+t)dt1/(1+t)=∑(-t)^k,(k:0→∞)∫lnt/(1+t)dt=∑∫lnt(-t)^kdt∫lnt*t^kdt=1/(k+1)(t^(k+1)*lnt-