∫dx/x(x^6+4)=?

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∫dx/x(x^6+4)=?

∫dx/x(x^6+4)=?
∫dx/x(x^6+4)=?

∫dx/x(x^6+4)=?
∫dx/[x(x^6+4)]
=∫x^2dx/[x^3*(x^6+4)]
令x^3=t--->dt=3x^2*dx--->x^2*dx=dt/3
因此原式=(1/3)∫dt/[t(t^2+4)
化有理分式1/[t(t^2+4)]为部分分式(1/4)/x-(t/4)/(t^2+4)
故前式=(1/24)∫[2dt/t-2tdt/(t^2+4)]
=(1/24)*2ln|t|-(1/24)ln(t^2+1)+C
=(1/24)ln[t^2/(t^2+4)]+C
=(1/24ln[x^6/(x^6+4)]+C.

估计要用换元法..令 x^3=t
然后套个啥公式之类的..应该是三角函数的...

设x=1/t∫1/x(x^6+4)dx=积分(1/1/t(1/t^6+4)d1/t=积分(t^7*(-1/t^2)/(1+4t^6)dt=-积分t^5/(1+4t^6)dt=-1/6积分1/(1+4t^6)d(1+4t^6)=-1/6ln|1+4t^6|+C
最后将t=1/x代入

∫dx/x(x^6+4)
=∫x^5dx/[x^6(x^6+4)]
=(1/6)∫d(x^6)/[x^6(x^6+4)]
=(1/24)∫(x^6+4-x^6)d(x^6)/[x^6(x^6+4)]
=(1/24)∫[1/x^6-1/(x^6+4)]d(x^6)
=(1/24)∫d(x^6)/x^6-(1/24)∫d(x^6+4)/(x^6+4)
=[lnx^6-ln(x^6+4)]/24+C
=ln[x^6/(x^6+4)]/24+C

倒代换