求助一道高数题,会的帮下忙〜一定有好评!
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/08/19 00:34:07
求助一道高数题,会的帮下忙〜一定有好评!
![](http://img.wesiedu.com/upload/5/7e/57efee3cc52087cc58a9358e265441b1.jpg)
![](http://img.wesiedu.com/upload/5/7e/57efee3cc52087cc58a9358e265441b1.jpg)
![求助一道高数题,会的帮下忙〜一定有好评!](/uploads/image/z/3147084-36-4.jpg?t=%E6%B1%82%E5%8A%A9%E4%B8%80%E9%81%93%E9%AB%98%E6%95%B0%E9%A2%98%2C%E4%BC%9A%E7%9A%84%E5%B8%AE%E4%B8%8B%E5%BF%99%26%2312316%3B%E4%B8%80%E5%AE%9A%E6%9C%89%E5%A5%BD%E8%AF%84%21)
∫dx/[x^(1/2)+x^(1/3)],u=x^1/6,x=u^6,dx=6u^5du
= 6∫u^5du/(u³+u²)
= 6∫u³du/(u+1)
= 6∫[u²-u-1/(u+1)+1]du
= 6[u³-u²-ln|u+1|+u] + c
= 2x^(1/2)-3x^(1/3)+6x^(1/6)-6ln|1+x^(1/6)| + c 再答: ���Dz�����ֵ���
再答: ����־�ֻҪ�� = 6[u³-u²-ln|u+1|+u] + c ��һ����u�ش��ٴ�����ݾ�����
再答: Ŷ�����ˣ�Ҫȥ���Ǹ�C
再问: лл��
再答: ������
= 6∫u^5du/(u³+u²)
= 6∫u³du/(u+1)
= 6∫[u²-u-1/(u+1)+1]du
= 6[u³-u²-ln|u+1|+u] + c
= 2x^(1/2)-3x^(1/3)+6x^(1/6)-6ln|1+x^(1/6)| + c 再答: ���Dz�����ֵ���
再答: ����־�ֻҪ�� = 6[u³-u²-ln|u+1|+u] + c ��һ����u�ش��ٴ�����ݾ�����
再答: Ŷ�����ˣ�Ҫȥ���Ǹ�C
再问: лл��
再答: ������