x^2arctanx/1+x^2
求不定积分(x2arctanx)/1 x2dx -
即可把原积分化成有理函数的积分,结果是\r\n-(arctanx)\\/x + ln(x绝对值) - 1\\/2 * ln(1+x^2)\r\n第二题把1\\/x^2*(x+1)写成1\\/(x+1) + (1-x)\\/x^2即可,
lim(x->0) x^2 arctan(1/x^2) =0= f(0)
即可把原积分化成有理函数的积分,结果是\r\n-(arctanx)\\/x + ln(x绝对值) - 1\\/2 * ln(1+x^2)\r\n第二题把1\\/x^2*(x+1)写成1\\/(x+1) + (1-x)\\/x^2即可,
lim(x->0) x^2 arctan(1/x^2) =0= f(0)
解答过程如下:
x^2arctan(1/x)~x 1-cos√x~(1/2)x 上述无穷小等价关系看出,分子和分母都是x的1阶式所以:lim(e^x-1)/(1-√cosx) + (x^2arctan1/x)/(1-√cosx)=2+2=4
求不定积分∫x^2arctanx/1+x^2dx -
简单计算一下即可,答案如图所示,
简单计算一下即可,答案如图所示,
分部积分[x^2arctanx/(1+x^2)]dx -
∫x^2arctanx/(1+x^2)dx =∫(x^2+1-1)arctanx/(1+x^2)dx =∫arctanxdx-∫arctanx/(1+x^2)dx =xarctanx-∫ x/(1+x^2)dx-∫arctanxd (arctanx)=xarctanx-(1/2)∫ 1/(1+x^2)d(x^2)-(1/2)(arctanx)^2 =xarctanx-(1/2)ln(1+x^2)-(1/2)(arctan希望你能满意。
简单计算一下即可,答案如图所示,
求x^2arctan(1/x)的极限,x趋向与0 -
e^x-1~x x^2arctan(1/x)~x 1-cos√x~(1/2)x 上述无穷小等价关系看出,分子和分母都是x的1阶式所以:lim(e^x-1)/(1-√cosx) + (x^2arctan1/x)/(1-√cosx)=2+2=4
设:arctanx²=α,则:tanα=1/x²设:arctan(1/x²)=β,则:tanβ=1/x²则:tanαtanβ=1 (sinα/cosα)×(sinβ/cosβ)=1 cosαcosβ-sinαsinβ=0 cos(α+β)=0 因为:0<α,β<π/2,则:α+β=π/2 即:arctan(x²)+arctan(1/x等我继续说。
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x^2arctanx/1+x^2