(x-3)根号4-x²dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/04/29 08:35:37
∫(x+√2-(√2+4)/(x+√2)dx=∫1dx-(√2+4)∫1/(x+√2)=x-(√2+4)ln|x+√2|+C
∫dx/√(4x-x^2)=∫dx/√([4-(x-2)^2]=arcsin[(x-2)/2]+C
$x[(x)^(1/2)+1)]dx=$[(x^(3/2)+x]dx=(5/2)*x^(5/2)+x^2/2(积分号9到4)=(5/2)*(9)^(5/2)+(9)^2/2-(5/2)*(4)^(5/
∫[3^(1/x)/x²]dx=-∫3^tdt……t=1/x=-(3^t)/ln3+C=-[3^(1/x)]/ln3+C;∫√x/(√x-1)dx=∫[1+1/(√x-1)]dx=x+∫2√
根号下(sinx-(sinx)^3)dx=根号下(sinx[1-(sinx)^2])dx=根号下(sinx*cos^2x)dx=根号下(sinx)*cosxdx=根号下(sinx)*dsinx=2/3
∫√[1+√x]/x^[3/4]dxLetu=x,dx=4udu=∫√[1+u]/u*[4u]du=4∫√[1+u]duLetu=tanz,du=seczdz=4∫√[1+tanz][seczdz]=
看得到图吗?积分上下限可改,另一个没看懂.
令x=2sinu,则:sinu=x/2,u=arcsin(x/2),dx=(1/2)cosudu.∴∫[√(4-x^2)/x^2]dx=∫[cosu/(sinu)^2]cosudu=∫[(cosu)^
换元,凑微分过程如下图:
(x^2)/2-18x^(1/2)+3x+C0.5*x^2+2*x^(1/2)+C9x-2x^3+0.2*x^5+C
∫1/[1+(√3x)]dx=1/√3·∫1/[1+(√3x)]d(√3x)=1/√3·∫1/[1+(√3x)]d(1+√3x)=1/√3·ln|1+√3x|+C
dx^(1/2)=(1/2)x^(-1/2)dx∫x^(-1/2)lnxdx=2∫lnxdx^(1/2)
解∫x√(4x²-1)dx=1/8∫√(4x²-1)d(4x²-1)=1/8∫√udu=1/8×(2/3)×u^(3/2)+C=1/12(4x²-1)^(3/2
设x=t的6次方∴t=6次根号下t∫1/(t³+t²)dt的6次方=∫6t的5次方/t²(t+1)dt=∫6t³/(t+1)dt=6∫(t³+1-1)
∫dx/(根号5-4x-x^2)=积分1/根号(3^2-(x+2)^2)d(x+2)=1/3积分1/根号(1-[(x+2)/3]^2)d(x+2)=积分1/根号(1-[(x+2)/3]^2)d[(x+
1.∫_(-1)^(2)1/(11+5x)³dx=(1/5)∫_(-1)^(2)1/(11+5x)³d(5x)=(1/5)∫_(-1)^(2)(11+5x)^(-3)d(11+5x
分步积分法原式=xarctan√x-∫xdarctan√x=xarctan√x-∫x/(1+x)dx=xarctan√x-∫(x+1-1)/(1+x)dx=xarctan√x-∫[1-1/(1+x)]