1/x(x^6+4)的不定积分怎么求
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1/x(x^6+4)的不定积分怎么求
![1/x(x^6+4)的不定积分怎么求](/uploads/image/z/18527612-68-2.jpg?t=1%2Fx%28x%5E6%2B4%29%E7%9A%84%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E6%80%8E%E4%B9%88%E6%B1%82)
因为d(x^3)=3x^2dx
所以dx=d(x^3)/3(x^2)
原式=∫1/[(x^6+4)*x*3x^2] dx
=∫1/[(x^6+4)*3x^3] d(x^3)
令x^3=t
原式=(1/3)∫1/[(t^2+4)t] d(t)
=(1/3)∫1/[(t^2+4)t*2t] d(t^2)
=(1/6)∫1/[(t^2+4)t^2] d(t^2)
=(1/24)∫(1/t^2)-[1/(t^2+4)] d(t^2)
=(1/24)∫(1/t^2)d(t^2)-(1/24)∫1/(t^2+4)d(t^2)
=(lnt^2)/24-[ln(t^2+4)]/24
所以原式=(lnx^6)/24-[ln(x^6+4)]/24
=(lnx)/4-[ln(x^6+4)]/24
所以dx=d(x^3)/3(x^2)
原式=∫1/[(x^6+4)*x*3x^2] dx
=∫1/[(x^6+4)*3x^3] d(x^3)
令x^3=t
原式=(1/3)∫1/[(t^2+4)t] d(t)
=(1/3)∫1/[(t^2+4)t*2t] d(t^2)
=(1/6)∫1/[(t^2+4)t^2] d(t^2)
=(1/24)∫(1/t^2)-[1/(t^2+4)] d(t^2)
=(1/24)∫(1/t^2)d(t^2)-(1/24)∫1/(t^2+4)d(t^2)
=(lnt^2)/24-[ln(t^2+4)]/24
所以原式=(lnx^6)/24-[ln(x^6+4)]/24
=(lnx)/4-[ln(x^6+4)]/24