∫dx(x^2-6x+5)

∫x/(x^2+5x+6)dx∫x/(x+2)(x+3)dx=∫x×[1/(x+2)-1/(x+3)]dx=∫x/(x+2)-x/(x+3)dx=∫1-2/(x+2)-1+3/(x+3)dx=∫3/(x+3)-2/(x+2)dx=3ln|x

∫(√(x^2+6x))dx∫(√(x²+6x+9-9))dx=∫(√（（x+3）²-9))dx=∫(√（（x+3）²-3²))dx换：t=x+3=∫(√（t²-3²))dt&nb

∫x^6/1+x^2dx

∫X√(2-5X)dx

∫x/（x^2+5）dx∫x/（x^2+5）dx=1/2(ln|x^2+5|)+C

求几个函数的不定积分∫(x^3-3^x)dx∫(6x^2-3x+5)dx∫(2x+3)dx∫[1/(2x+1)]dx∫[1/(4x-1)]dx∫e^(3x)dx∫(x^3-3^x)dx=3/4x^4-3^x/ln3+c∫(6x^2-3x+5

∫4/((2-6X-X^2)^(5/2))dx设I=∫4/(2-6x-x^2)^(5/2)dx=∫4/[11-(x+3)^2]^(5/2)dx令x+3=√11cost(0-4/121(tant+1/3tan^3t)+C对分母配方，然后用三角

∫[(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

∫1/(x^2+5x+6)dx∫1/(x^2+5x+6)dx=∫[x+3-(x+2)]/(x^2+5x+6)dx=∫[1/(x+2)-1/(x+3)]dx=∫1/(x+2)dx-∫[/(x+3)dx=ln(x+2)-ln(x+3)

求不定积分∫(1/x^2+6x+5)dx∫(1/x^2+6x+5)dx=∫(1/[(x+5)(x+1)]dx=1/4∫(1/(x+1)-1/(x+5)dx=1/4[ln(x+1)-ln(x+5)]+C非要用什么换元法的话,令x+1=t,dx

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)

求∫dx/[(1-x^2)^5/2]的不定积分还有∫x^2/(1-x^6)dx的不定积分回答晚了，请采纳那位网友的，他回答得很棒。

∫(3x-2)/(x^2-2x+5)dx

∫[（x+1)/(x^(2)+2x+5]dx原式=0.5∫d（x²+2x+5）/(x²+2x+5)=0.5㏑(x²+2x+5)

∫x/(x^4+2(x^2)+5)dx令a=x²原式=1/2*∫dx²/(x^4+2x²+5)=1/2*∫da/(a²+2a+1+4)=1/2*∫da/[(a+1)²+4]=1/8*∫da/

∫2x²+3x-5/x+3dx设x+3＝t→dx＝dt,代入原式得∫[(2x²+3x-5)/(x+3)]dx＝∫[(2(t-3)²+3(t-3)-5)/t]dt＝∫[2t+(4/t)-9]dt＝t²+

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

∫x/√(x^2+4x+5)dx∫x/√(x²+4x+5)dx=(1/2)∫(2x+4-4)/√(x²+4x+5)dx=(1/2)∫(2x+4)/√(x²+4x+5)dx-(1/2)(4)∫1/√(x²

∫1/(x(√x+x^(2/5)))dx设x=t^10,则dx=10t^9*dt∴原式=∫10t^9*dt/[t^10(t^5+t^4)]=∫[1/t-1/t²+1/t³-1/t^4+1/t^5-1/(t+1)]dt=l

∫x/((x^2)+4x+5)dx使用换元法就可以解本题了.