∫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)]
=∫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

∫dx/x(x^6+4)=? ∫cos x / ( 1 + (sinx)^2 ) dx = ∫x^3 / ( 1 + x^4 ) dx = ∫(sec x)^3 * tan x dx = ∫x^2 * e^(-2x) dx = ∫x * cos 2x dx = ∫(cos 2x)^2 dx = ∫(13x - 6) / ( x (x-2)(x+3) )dx = ∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx = 请把过程写出来哈.>< 旷这个怎恶魔算∫cos x / ( 1 + (sinx)^2 ) dx = ∫x^3 / ( 1 + x^4 ) dx = ∫(sec x)^3 * tan x dx = ∫x^2 * e^(-2x) dx = ∫x * cos 2x dx = ∫(cos 2x)^2 dx = ∫(13x - 6) / ( x (x-2)(x+3) )dx = ∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx = dx/(x+4)=? ∫1/(1-x^4)dx=? ∫f(x)dx+∫xf'(x)dx= d/dx∫(x,0)f(3x)dx= ∫ dx /4X+5 ∫ x cosx dx=? ∫cot^x dx= ∫(x+lnx)dx=? ∫xf'(x)dx=? ∫x^6/(1-x^4)dx不定积分 ∫4/((2-6X-X^2)^(5/2)) dx ∫1/x(x的6次方+4)dx ∫sin（6x）sin（4x）dx ∫sin(6x)sin(2x)dx=? ∫(1-x)/[√(9-4x^2)]dx=