dx=∫cost(1-(sint)^2)^3/2dt=∫(cost)^4dt=∫((cos4t)/8+(cos2t)/2+3/8)dt(二倍角公式得到的)=-(sin4t)/32-(sin2t)/4+3t/8
=-sintcost(1-2(sint)^2)/8-sintcost/2+3t/3(还是二倍角)=-x(1-x^2)^1/2(5-2x^2)/8+3arcsinx/8
∫1/(x^(1/2)+x^(1/3)) dx=∫1/(t^3+t^2) *6t^5 dt
=S6t^3/(t+1)dt
=6S(t^3+t^2-t^2-t+t+1-1)(t+1) dt
=6S(t^2-t+1)dt-6S1/(t+1)d(t+1)
=2t^3-3t^2+6t-6ln|t+1|+c
=2(x^1/2)-3(x^1/3)+6(x^1/6)-6ln(1+x^1/6)+C本回答被提问者采纳
=∫dx/[x^(1/3)(x^(1/6)+1)]
=(3/2)∫dx^(2/3)/[x^(1/6)+1]
=(3/2)∫4x^(3/6)dx^(1/6)/[x^(1/6)+1]
=6∫[x^(3/6)+x^(2/6)]dx^(1/6)/[x^(1/6)+1] -6∫[x^(2/6)+x(1/6)]dx^(1/6)/[x^(1/6)+1]
+6∫[x^(1/6)+1]dx/[x^(1/6)+1] -6∫d[x^(1/6)+1]/[x^(1/6)+1]
=6∫x^(2/6)dx^(1/6) - 6∫x^(1/6)dx^(1/6) +6∫dx^(1/6) -6ln|x^(1/6)+1|
=2x^(3/6)-3x^(2/6)+6x^(1/6)-6ln(x^(1/6)+1)+C
1\/(X^1\/2+X^1\/3)DX的不定积分 谁能帮我做下,
简单计算一下即可，答案如图所示
求不定积分∫1\/[x^(1\/2)-x^(2\/3)]dx求高手解题要步骤谢谢
原式=∫2x^(1\/2)\/[x^(1\/2)-x^(2\/3)]dx^(1\/2)令x^(1\/2)=u 原式=∫2u\/[u-u^\/3]du =∫2\/[1-u^\/2]du =2arctanu+c 将u=x^(1\/2)=代入上式 原式=2arctan[x^(1\/2)]+c
求1\/(x^1\/2+x^1\/3)的不定积分
简单计算一下即可，答案如图所示
那个1\/x^1\/2 x^1\/3求不定积分怎么求啊?
dx=6u^5 du u^3 =u^2.(u+1) - u^2 =u^2.(u+1) - u(u+1) +u =u^2.(u+1) - u(u+1) +(u+1) -1 ∫dx\/[x^(1\/2) +x^(1\/3)]=∫6u^5 du\/(u^3 +u^2)=6∫u^3 \/(u +1) du =6∫[ u^2 -u +1 - 1\/(u +1)] du =6[(1\/3)u^3 -(1\/...
求不定积分∫1\/(x^1\/3+x^1\/2)dx
简单计算一下即可，答案如图所示
求1\/(x^(1\/3)+x^2)的不定积分
这题应该是：S1\/（x^(1\/3)+x^(1\/2))dx t=x^(1\/6),x=t^6,dx=6t^5dt 上积分=S1\/(t^2+t^3)*6t^5dt =6S(t^3\/(t+1))dt =6S(t^3+t^2-t^2-t+t+1-1)\/(t+1)dt =6S(t^2-t+1-1\/(t+1))dt =2t^3-3t^2+6t-6ln(t+1)+c =2*x^(1\/2)-3*x^(1...
求1╱(x∧1╱2+x^1\/4)dx不定积分
设 x = t^4，则 dx = 4*t^3*dt 积分公式转化成：=∫4*t^3*dt\/(t^2 + t)=4*∫t^2\/(t+1) *dt =4*∫[(t^2-1) +1]\/(t+1) *dt =4*[∫(t-1)dt + ∫dt\/(t+1)]=4*[∫t*dt -∫dt + ln|t+1|]=4*[1\/2*t^2 - t + ln|t+1|] +C =2*t^2 - ...
x^(1\/3)\/x(x^1\/2 +x^1\/3)求不定积分
点击放大，荧屏放大再放大：
∫(dx\/((1+x^1\/3)x^1\/2))计算不定积分
设x=t^6, t>0, 对原式做代换：∫(dx\/((1+x^1\/3)x^1\/2))=∫1\/t^3(1+t^2) d t^6 =6∫t^2\/(1+t^2) dt =6∫[1- 1\/(1+t^2)]dt =6(t-arctant)+c 根据 t=x^(1\/6)回代：=6[x^(1\/6)-arctanx^(1\/6)]+c 以上答案仅供参考，如有疑问可继续追问。
求1\/(x^2+x^3)的不定积分
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