不定积分∫ln(1+x^2)dx
不定积分∫ln(1+x^2)dx
原式=xln(1+x^2)-∫xd[ln(1+x^2)] 这些步骤怎么变的xln(1+x^2)怎么就到∫外了那个x怎么来的希望能讲一下原理课本上一步就到了看不懂怎么回事
=xln(1+x^2)-∫2x^2/(1+x^2)dx
=xln(1+x^2)-2∫[1-1/(1+x^2)dx
=xln(1+x^2)-2x+2acrtgx+C
原式=xln(1+x^2)-∫xd[ln(1+x^2)] 这些步骤怎么变的xln(1+x^2)怎么就到∫外了那个x怎么来的希望能讲一下原理课本上一步就到了看不懂怎么回事
=xln(1+x^2)-∫2x^2/(1+x^2)dx
=xln(1+x^2)-2∫[1-1/(1+x^2)dx
=xln(1+x^2)-2x+2acrtgx+C
数学人气:859 ℃时间:2020-01-25 06:07:20
优质解答
用分部积分法,(uv)'=u'v+uv',设u=ln(1+x^2),v'=1,u'=2x/(1+x^2),v=x,原式=xln(1+x^2)-2∫x^2dx/(1+x^2)=xln(1+x^2)-2∫(1+x^2-1)dx(1+x^2)=xln(1+x^2)-2∫dx+2∫dx/(1+x^2)=xln(1+x^2)-2x+2arctanx+C.用分部积分法,(uv)'=u'v+uv',设u=ln(1+x^2),v'=1,u'=2x/(1+x^2),v=x,和这一步原式=xln(1+x^2)-2∫x^2dx/(1+x^2)有什么联系吗恕我愚昧没看懂我明白了两边同时积分对否积分结果=uv-∫u'vdx,uv=xln(1+x^2),∫u'vdx=∫2x*xdx/(1+x^2),再把分子中x^2配一个+1,约掉分母。
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