求不定积分∫1 (x^2-2x 3)dx-搜答案-智慧作业帮

=-1/2∫√(1-x^2)d(1-x^2)=-1/2×2/3√(1-x^2)^3+C=-1/3√(1-x^2)^3+C

差不多就这样再答：

∫x/(1+x^2)dx=1/2∫1/(1+X^2)d(x^2)=1/2∫1/(1+X^2)d(1+x^2)=1/2ln(1+x^2)+c

∫(arctanx)/(x^2(x^2+1))dx=∫(arctanx)/x^2dx-∫(arctanx)/(x^2+1)dx=∫(arctanx)d(1/x)-∫(arctanx)darctanx=

∫1/(x^2+2x+5)dx=∫1/[(x+1)^2+4]dx=∫(1/4)/[[(x+1)/2]^2+1]dx=∫(1/4)·2/[[(x+1)/2]^2+1]d((x+1)/2)=(1/2)∫1

∫(x^2-1)sin2xdx先括号拆开=∫x^2*sin2xdx-∫sin2xdx=-1/2*∫x^2dcos2x-1/2*∫sin2xd2x先凑微分=-1/2*∫x^2dcos2x-1/2*∫si

我的解答如下：换元法令x=3/2sint,t∈[-0.5π,0.5π]dx=3/2cost带入后得到∫(1-x)/[√(9-4x^2)]dx=∫(1-1.5sint)1.5costdt/3cost=∫

换元法,令x=rcostdx=-rsintdt代入即可

1、原式=∫(sinx^2)*(cosx^2)dsinx=∫sin^2*(1-sinx^2)dsinx=∫(sin^2-sinx^4)dsinx=∫sinx^2dsinx-∫sinx^4dsinx=1

原式=∫(x+1)/x²+∫xlnxdx=∫x/x²+∫1/x²+1/2∫lnxdx²=∫1/x+∫1/x²+1/2*x²lnx-1/2∫x

很简单啊,好好观察形状就好解了

分部积分法∫xe^x/(1+x)^2dx＝－∫xe^xd[1/(1+x)]＝－xe^x/(1+x)＋∫(1+x)e^x×1/(1+x)dx＝－xe^x/(1+x)＋∫e^xdx＝－xe^x/(1+x)

∫(x^2-3x)/(x+1)dx=∫[(x+1)(x-4)/(x+1)+4/(x+1)]dx=∫(x-4)dx+∫4/(x+1)dx=x²/2-4x+4ln(x+1)+C其中C为任意常数

积分:(x^2+1)/(x^4+1)dx=积分:(1+1/x^2)/(x^2+1/x^2)dx(上下同时除以x^2)=积分:d(x-1/x)/[(x-1/x)^2+(根号2)^2]=1/根号2*arc

解∫x/(x^2)dx=∫1/xdx=ln|x|+C

∫(2x+1)dx=∫2xdx+∫dx=x^2+x+C