原式
=∫[(x+2)/[2(x+1)^2]-x/[2(x^2+1)] dx
=1/2*∫[(x+1+1)/(x+1)^2-x/(x^2+1) dx
=1/2*∫(1/(x+1)+1/(x+1)^2-x/(x^2+1))dx
=1/2*[ln|x+1|-1/(x+1)-1/2*ln(x^2+1)]+C
=1/2*ln|(x+1)/√(x^2+1)|-1/2(x+1)+C本回答被提问者采纳
求下列不定积分。 (1)∫[1\/(x+1)^2 (x^2+1)]dx (2) ∫[1\/(2+sinx...
=1\/2*∫1\/(1+1\/2*sinx)dx 令t=tan(x\/2),则sinx=2t\/(1+t^2),cosx=(1-t^2)\/(1+t^2),dx=2dt\/(1+t^2)=∫dt\/(1+t+t^2)=4\/3*∫1\/{[(2t+1)\/√3]^2+1} =4\/3*√3\/2*arctan[(2t+1)\/√3]+C =2\/√3*arctan[(2tan(x\/2)+1)\/√3]+C 上面这个结果可以...
求∫1\/[(x+1)^2(x^2+1)]dx的不定积分,
简单分析一下，答案如图所示
∫1\/[(x+1)(x^2+x+1)^2]dx求不定积分
这是一个有理函数的积分，这种积分没什么技巧，过程烦。先分成部分分式，然后进行积分。这里给个原函数，自己试试。log就是ln.tan的-1次方表示arctan
∫1\/[x^2(x^2+1)]dx的不定积分怎么算
=∫[1\/x^2 -1\/(x^2+1)] dx = -1\/x - arctanx + C
求∫1\/[x(x^2+1)]dx的不定积分
法一：令u=x^2,则du=2xdx ∫1\/[x(x^2+1)]dx =1\/2·∫1\/[u(u+1)]du =1\/2·∫[1\/u-1\/(u+1)]du =1\/2·∫1\/u du-1\/2·∫1\/(u+1) d(u+1)=1\/2·lnu-1\/2·ln(u+1)+C =1\/2·ln[u\/(u+1)]+C =1\/2·ln[x^2\/(x^2+1)]+C 法二：∫1\/[x(x^2...
求(x+1)^2\/x^1\/2不的定积分
求（x+1）\/(x^2+1)^2的不定积分 ∫[(x+1)\/(x^2+1)^2]dx 令x=tant，则：dx=d(tant)=sec^2 tdt 原积分=∫[(tant+1)\/sec^4 t]*sec^2 tdt =∫[(tant+1)\/sec^2 t]dt =∫{[(sint\/cost)+1]\/(1\/cos^2 t)}dt =∫(sintcost+cos^2 t)dt =∫sintcostdt+∫cos^2...
∫(1\/( x^2+1)^2) dx的不定积分为多少?
∫（1\/（x^2+1）^2）dx的不定积分为1\/2*x\/(1+x^2)+1\/2arctanx+C。解：令x=tant，则t=arctanx，且x^2+1=(tant)^2+1=(sect)^2 ∫（1\/（x^2+1）^2）dx =∫（1\/(sect)^4）dtant =∫（(sect)^2\/(sect)^4）dt =∫（1\/(sect)^2）dt =∫(cost)^2dt =1\/2∫(...
求不定积分∫[1\/(x+1)(x²+1)]dx
答案如下图：
不定积分(x+1)^2\/(x^2+1)^2
求（x+1）\/(x^2+1)^2的不定积分 ∫[(x+1)\/(x^2+1)^2]dx 令x=tant，则：dx=d(tant)=sec^2 tdt 原积分=∫[(tant+1)\/sec^4 t]*sec^2 tdt =∫[(tant+1)\/sec^2 t]dt =∫{[(sint\/cost)+1]\/(1\/cos^2 t)}dt =∫(sintcost+cos^2 t)dt =∫sintcostdt+∫cos^2...
1\/(x^2+1)^2的不定积分怎么算
∫（1\/（x^2+1）^2）dx的不定积分为1\/2*x\/(1+x^2)+1\/2arctanx+C。解：令x=tant，则t=arctanx，且x^2+1=(tant)^2+1=(sect)^2 ∫（1\/（x^2+1）^2）dx =∫（1\/(sect)^4）dtant =∫（(sect)^2\/(sect)^4）dt =∫（1\/(sect)^2）dt =∫(cost)^2dt =1\/2∫(...
