求不定积分∫2x·1\/1+x^2dx
回答：x=tan y, I=Integral[sin2y]=-cos2y\/2=-(1-x^2)\/(1+x^2)\/2+C
∫x^2\/1+x^2dx的不定积分怎么算
=∫(1+x^2-1)\/(1+x^2)dx =∫1-1\/(1+x^2)dx =x-arctanx+C
求解不定积分∫√(x^2+2x)\/x^2dx,要过程
2015-04-24 求不定积分∫x^2\/√a^2+x^2dx,谢谢~ 1 2012-11-30 求不定积分:∫x^2dx\/根号(a^2-x^2)= 55 2016-12-18 高数 求解不定积分 ∫dx\/(1-x)^(5\/2) 1 2017-04-19 求不定积分∫√(R^2-x^2)x^5 dx 3 2012-04-12 求不定积分∫2dx\/x√a^2-x^2 1 2017-01-03 ...
求fx^2\/1+x^2dx的不定积分
2015-05-02 计算不定积分(x^2)\/(1+x^2)^2dx 27 2015-12-14 ∫x^2\/1+x^2dx的不定积分怎么算 2016-03-12 ∫x^2\/1+x^2dx的不定积分怎么算 3 2015-12-15 求(-x^2-2)\/(x^2+x+1)^2dx的不定积分,用... 4 2016-02-03 (x^2+k^2)^1\/2dx求不定积分 2 2018-01-09 √(1...
不定积分∫x^2\/(1+x^2)^2dx
∫ x²\/(1 + x²)² dx,令x = tanz,dx = sec²z dz = ∫ tan²z\/sec⁴z * (sec²z dz)= ∫ sin²z\/cos²z * cos²z dz = ∫ (1 - cos2z)\/2 dz = z\/2 - (1\/4)sin2z + C = (1\/2)arctanx - (1\/2) ...
∫tan^2x\/1+x^2dx求不定积分
令u=tanx,则du=dx\/(1+x^2).∫tan^2x\/1+x^2dx =∫u^2 du = u^3\/3 + C = tan^3 x + C
定积分(2到1)(x^2+1\/x^2)dx
先求不定积分 =∫x^2+1\/x^2dx =x^3\/3-1\/x+C 代值进去 =8\/3-1\/2-1\/3+1 =17\/6
定积分(2到1)(x^2+1\/x^2)dx
先求不定积分 =∫x^2+1\/x^2dx =x^3\/3-1\/x+C 代值进去 =8\/3-1\/2-1\/3+1 =17\/6
不定积分∫x^2\/(x^2+1)^2dx怎么求?
方法如下，请作参考：若有帮助，请采纳。
不定积分∫1\/(x^2(1+x^2))dx, 求解
∫1\/(x^2(1+x^2))dx = ∫[(1+x^2)-x^2]\/(x^2(1+x^2))dx = ∫[1\/x^2-1\/(1+x^2)]dx =∫1\/x^2dx - ∫1\/(1+x^2)dx = -1\/x -arctan x + C
