x/(x的平方+2x+3)dx的不定积分是多少?
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x/(x的平方+2x+3)dx的不定积分是多少?
![x/(x的平方+2x+3)dx的不定积分是多少?](/uploads/image/z/17293459-67-9.jpg?t=x%2F%28x%E7%9A%84%E5%B9%B3%E6%96%B9%2B2x%2B3%29dx%E7%9A%84%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E6%98%AF%E5%A4%9A%E5%B0%91%3F)
∫x/(x的平方+2x+3)dx
=∫x/[(x+1)²+2] dx
=∫[(x+1)-1]/[(x+1)²+2] dx
=∫[(x+1)/[(x+1)²+2] dx-∫1/[(x+1)²+2] dx
=1/2∫1/(x的平方+2x+3) d(x的平方+2x+3) -1/√2 arctan(x+1)/√2+c
=1/2ln(x的平方+2x+3)-√2/2 arctan(x+1)/√2+c
再问: 1/2∫1/(x的平方+2x+3) d(x的平方+2x+3) -1/√2 arctan(x+1)/√2+c 这步怎么来的啊,请详解
=∫x/[(x+1)²+2] dx
=∫[(x+1)-1]/[(x+1)²+2] dx
=∫[(x+1)/[(x+1)²+2] dx-∫1/[(x+1)²+2] dx
=1/2∫1/(x的平方+2x+3) d(x的平方+2x+3) -1/√2 arctan(x+1)/√2+c
=1/2ln(x的平方+2x+3)-√2/2 arctan(x+1)/√2+c
再问: 1/2∫1/(x的平方+2x+3) d(x的平方+2x+3) -1/√2 arctan(x+1)/√2+c 这步怎么来的啊,请详解