Integrations by substitution
回答 (2)
2009-08-11 8:14 pm
2009-08-11 12:27 pm
u=1/x
du/dx=1/x^2
du=1/x^2 dx
(1-x^2)/x^4
=1/x^4-x^2/x^4
=1/x^4-1/x^2
1/x^4-1/x^2 dx
=(1/x^2-1)(1/x^2) dx
=u^2-1 du
=(1/3)u^3-u
(1/3)(1/x)^3-1/x
=(1/3)(1/x^3)-1/x
=1/3x^3-1/x
du/dx=1/x^2
du=1/x^2 dx
(1-x^2)/x^4
=1/x^4-x^2/x^4
=1/x^4-1/x^2
1/x^4-1/x^2 dx
=(1/x^2-1)(1/x^2) dx
=u^2-1 du
=(1/3)u^3-u
(1/3)(1/x)^3-1/x
=(1/3)(1/x^3)-1/x
=1/3x^3-1/x
收錄日期: 2021-04-16 16:20:02
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