求一不定积分求上图中的不定积分,
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求一不定积分
![](http://img.wesiedu.com/upload/9/b4/9b4e09477915bf2bccd6ffa75ad33c3b.jpg)
求上图中的不定积分,
![](http://img.wesiedu.com/upload/9/b4/9b4e09477915bf2bccd6ffa75ad33c3b.jpg)
求上图中的不定积分,
![求一不定积分求上图中的不定积分,](/uploads/image/z/18636470-62-0.jpg?t=%E6%B1%82%E4%B8%80%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E6%B1%82%E4%B8%8A%E5%9B%BE%E4%B8%AD%E7%9A%84%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%2C)
答:
∫ (2-u) / (2u-2u²+2u³) du
=∫ 1/[ u(u²-u+1)] du -(1/2)*∫ 1/(u²-u+1) du
=∫ 1/u - (u-1) /(u²-u+1) du -(1/2)*∫ 1/(u²-u+1) du
=ln | u| -∫ (u-1+1/2) /(u²-u+1) du
= ln |u| -∫ (u-1/2) / [ (u-1/2)²+3/4 ] du
=ln |u| -(1/2) ∫ 1/ [(u-1/2)²+3/4 ] d [ (u-1/2)²+3/4 ]
=ln |u| -(1/2) ln (u²-u+1)+C
=ln | u/√(u²-u+1) | +C
∫ (2-u) / (2u-2u²+2u³) du
=∫ 1/[ u(u²-u+1)] du -(1/2)*∫ 1/(u²-u+1) du
=∫ 1/u - (u-1) /(u²-u+1) du -(1/2)*∫ 1/(u²-u+1) du
=ln | u| -∫ (u-1+1/2) /(u²-u+1) du
= ln |u| -∫ (u-1/2) / [ (u-1/2)²+3/4 ] du
=ln |u| -(1/2) ∫ 1/ [(u-1/2)²+3/4 ] d [ (u-1/2)²+3/4 ]
=ln |u| -(1/2) ln (u²-u+1)+C
=ln | u/√(u²-u+1) | +C