∫x(2x+3)^2dx

∫2^x-3dx令u=x-3,du=dx∫2^(x-3)dx=∫2^udu=2^u/ln2+C=2^(x-3)/ln2+C公式：∫a^xdx=a^x/lna+C

求助∫d/dx[X^tan(x^2)]dx和∫dx/(2+3X^2)∫d/dx[X^tan(x^2)]dx=∫d[X^tan(x^2)]=X^tan(x^2)+C∫dx/(2+3X^2)=(1/2)*∫dx/(1+(√(3/2)x)^2)=

∫dx/x(2x+3)^2

∫(2^x+3^x)²dx展开得到原积分=∫4^x+2*6^x+9^xdx=4^x/ln4+2*6^x/ln6+9^x/ln9+C,C为常数

∫x^3/9+X^2dx.我想你的题应该是这样吧∫x³/(9+x²)dx=(1/2)∫x²/(9+x²)d(x²)=(1/2)∫(x²+9-9)/(9+x²)d(x

∫X^2e^-X^3dx.原式=-1/3∫e^-X^3d(-X^3)=-1/3e^-X^3+c∫X^2e^(-X^3)dx=-1/3∫e^(-X^3)dx^3=-1/2e^(-X^3)+C

∫（x-3x+2）dx掉了一个2次方吧!∫（x^2-3x+2）dx=(1/3)x^3-(1/2)3x^2+2x+C=(1/3)x^3-(3/2)x^2+2x+C

∫x^3/1+x^2dx∫x^3/(1+x^2)dx=∫[x^3+x-x]/(1+x^2)dx=∫x-x/(1+x^2)dx=x²/2-1/2ln[1+x^2]+c你的好评是我前进的动力.我在沙漠中喝着可口可乐,唱着卡拉ok,骑着

∫(x-1)^2/x^3dx∫(x²-2x+1)/x³dx=∫(1/x-2/x²+1/x³)dx=lnx+2/x-2/x²+C

∫3+x/(9-x^2)dx根据公式来做就可以了，注意x²要重复积分。

∫x^3/(9+x^2)dx∫x^3/(9+x^2)dx=1/2∫x^2/(9+x^2)dx^2(x^2=t)=1/2∫t/(9+t)dt=1/2∫(t+9-9)/(9+t)dt=1/2∫[1-9/(9+t)]dt=1/2t-9/2ln(9

∫dx/(x-2)平方(x-3)∫{1/[(x-2)^2(x-3)]}dx1/[(x-2)^2(x-3)]=A/(x-2)+B/(x-2)^2+C/(x-3)=>1=A(x-2)(x-3)+B(x-3)+C(x-2)^2coef.ofx^2

∫(X^3)/(1+X^2)dx具体见图片内容：

∫x^2/x+3dxx^2/(x+3)=-(5/2)+9Log[4]-(9Log[9])/2

∫x^3*e^x^2dx原式=1/2∫x²e^x²dx²=1/2∫x²de^x²=1/2*x²e^x²-1/2∫e^x²dx²=1/2*x²

∫[(x^2-x+6)/(x^3+3x)]dx(x^2-x+6)/(x^3+3x)=2/x-(x+1)/(x^2+3).原式=∫2/xdx-∫(x+1)/(x^2+3)dx=2ln|x|-(1/2)ln(x^2+3)-(1/√3)arcta

x-9/[(根号)x]+3dx∫x+1/[(根号)x]dx∫[（3-x^2）]^2dx(x^2)/2-18x^(1/2)+3x+C0.5*x^2+2*x^(1/2)+C9x-2x^3+0.2*x^5+C看图！2和4两题。

求不定积分1.∫x√xdx2.∫x^2√xdx3.∫dx/x^24.∫6x^3dx√xdx表示根号xdx1.原式=∫x^(3/2)dx=2/5x^(5/2)+C2.原式=∫x^(5/2)dx=2/7x^(7/2)+C3.原式=∫x^(-2)

∫x^3/(x^8-2)dx∫(x^3-1)/(x^2+1)dx1.令t=x^4dt\4=x^3dx原式=∫dt\4(t^2-2)=2^(1\2)\16∫{dt\[t-2^(1\2)]-dt\[t+2^(1\2)]}=2^(1\2)\16I

∫(x^3-x^2+x+1)/(x^2+1)dx∫(x+4)/(x^2-x-2)dx1.先化为x-1+2/(x^2+1)再积分,=x^2乘以1/2-x+2乘以arctanx+c2.先化为(x+4)/（x-2)（x+1)=A/(x-2)+B/