(3X^2y 8Xy^2)dx (x^3 8x^2 12y^2)dy=0

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(3X^2y 8Xy^2)dx (x^3 8x^2 12y^2)dy=0
∫(2^x)/((2^x)+3)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

∫[(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)arctan(x

∫(1-x)^2/x^3 dx

∫(1-x)^2/x^3dx=∫(1-2x-x^2)/x^3dx=∫(x^(-3)-2x^(-2)+x^(-1))dx=1/(-3+1)x^(-3+1)-1/(-2+1)x^(-2+1)+ln|x|+

∫2/(x-3)dx怎么算?

∫[2/(x-3)]dx=2ln(x-3)+C再问:能不能写一两步过程再答:∫(1/x)dx=lnx+C则:∫[2/(x-3)]dx=2∫[1/(x-3)]d(x-3)=2/(x-3)+C

∫(x-1)^2/x^3 dx

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

cos(x^2)dx

再答:见图

∫x^3/1+x^2 dx

∫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

积分(3,1)dx/(x-2)=

∫(1->3)dx/(x-2)=[ln|x-2|](1->3)=ln1-ln1=0

不定积分x/(2-3x^2)dx

∫x/(2-3x^2)dx=1/2∫1/(2-3x^2)d(x^2)=-1/6∫1/(2-3x^2)d(2-3x^2)=-1/6*ln|2-3x^2|+C

∫dx/(2x-3)²

1/2∫1/(2x-3)²d(2x-3)=1/2×(-1/3)×1/(2x-3)³+C=-1/[6(2x-3)³]+C再问:题打错了是∫dx/[1+(2x+3)]再答:有

∫2^x*3^x/(9^x-4^x) dx

∫2^x*3^x/(9^x-4^x)dx=∫(2/3)^xdx/[1-(4/9)^x]=[ln(2/3)]^(-1)∫d[(2/3)^x]/{1-[(2/3)^x]^2}={[ln(2/3)]^(-1

x-9/[(根号)x]+3 dx ∫ x+1/[(根号)x] dx ∫ [(3-x^2)]^2 dx

(x^2)/2-18x^(1/2)+3x+C0.5*x^2+2*x^(1/2)+C9x-2x^3+0.2*x^5+C

∫dx/[(1+x^2)^(3/2)]

令x=tant,-π/2

∫(arctanx)^3/(1+x^2)dx

答:∫(arctanx)^3/(1+x^2)dx=∫(arctanx)^3d(arctanx)=(1/4)(arctanx)^4+C

∫(X^3)/(1+X^2)dx

具体见图片内容:再问:第二步怎么来的?没认真听课现在看起来很吃力麻烦讲解下我会提高悬赏的再答:就是自然对数lnx求导的形式:(lnx)'=1/x

∫X^2 e^-X^3 dx.

原式=-1/3∫e^-X^3d(-X^3)=-1/3e^-X^3+c

∫x^3/9+X^2 dx.

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

∫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

∫(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为常数再问:(⊙o⊙)哦看懂了谢谢再答:不必客气的啊~