不定积分x^7 (1 x^4)^2 5-搜答案-搜搜做题作业网

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1/(x^4-x^2)=-1/x^2-1/[2(x+1)]+1/[2(x-1)]积分=1/x+(1/2)ln(1-x)-(1/2)ln(1+x)+C如果要写短些1/x-arctanhx+C(保证是对的

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

∫x^7dx/(x^4+2)=(1/4)∫x^4d(x^4)/(x^4+2)=(1/4)x^4-(1/4)ln(x^4+2)+C∫（3x^4+x^3+4x^2+1)dx/(x^5+2x^3+x)=∫(

请问题目是x^2√(1+x^4)还是x^2/(1+x^4)再问：是根号再答：感觉你问的不能用初等函数函数表示，你确定题目是这样的，还是或许不用求原函数

令1/[(x-1)(x²+4x+9)]=A/(x-1)+(Bx+C)/(x²+4x+9)==>1=A(x²+4x+9)+(Bx+C)(x-1)1=Ax²+4Ax

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

令x=tany∫(x^2/(1+x^4))dx=∫(tany^2/(1+tany^4))*(1/(cosy)^2)dy=∫(siny)^2/((siny)^4+(cosy)^4)dy=∫(1/2)(1

∫x^4/(1+x²)]dx=∫[(x^4-1)+1]/(1+x²)]dx=∫(x^4-1)/(1+x²)+∫1/(1+x²)dx=∫(x²+1)(x

答：∫(x^4-2)/(x²+1)dx=∫(x^4-1-1)/(x²+1)dx=∫(x²-1)-1/(x²+1)dx=x³/3-x-arctanx+C

原式=∫[(x-1)(x+4)+8]/(x-1)dx=∫[x+4)+8/(x-1)dx=x²/2+4x+8ln|x-1|+C

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

答：1.原式=∫1/[(x+1)^2+4]dx=1/4∫1/[((x+1)/2)^2+1]dx=1/2*arctan[(x+1)/2]+C2.原式=1/2∫1/x-x^6/(x^7+2)dx=1/2[

∫(x-1)/√(9-4x^2)dx=∫x/√(9-4x^2)dx-∫1/√(9-4x^2)dx=-1/8*∫1/√(9-4x^2)d(9-4x^2)-0.5*∫1/√[1-(2x/3)^2]d(2x