Q:

Differential Equations; Bernoulli equation

I am stuck on the following Bernoulli equation,
$\frac{dx}{dt}+\frac{1}{4t\ln(t)}$ $x(t)=\frac{t^3}{\ln(t)}$ $x(t)^5$
I have changed it to a first order linear equation
$\frac{dz}{dt}-\frac{1}{t\ln(t)}$ $z=\frac{-4t^3}{\ln(t)}$
Where $z=\frac{1}{x^4}$ and $\frac{dz}{dt}=-4\frac{dx}{dt}\frac{1}{x^5}$ 
I am now solving using an integrating factor of $\ln(t)$ however I am stuck here. I think I have made a mistake getting to this point as it appears to be unsolvable. 
If someone could find my mistake or suggest another way of solving that would be great 

A:

You are correct . Now find the integrating factor and continue.
you have  $\frac {dz}{dt } -\frac{1}{t\ln(t)}z = \frac{-4t^3}{\ln(t)}$
the integrating factor IF = $\large e^{\int\frac{-1}{t\ln(t)}\,dt}$
there let $\ln(t) =w\implies \frac1{t}\,dt = dw$
IF=$\large e^{\int\frac{-1}{w}\,dw}  = \large e^{-\ln(w)}=\large e^{-\ln(\ln(t))}= \large e^{\ln(\frac{1}{\ln(t)})}= \frac{1}{\ln(t)}$
the solution is given by ;
$z\cdot \frac{1}{\ln(t)}=\int\frac{-4t^3}{\ln(t)}\cdot \frac{1}{\ln(t)}\,dt$
Can you continue from here?
EDIT : see @Isham s answer if you want to know about the last part.

A:

$$x'+\frac{1}{4t\ln(t)}x=\frac{t^3}{\ln(t)}x^5$$
Substitute $z=\frac 1 {x^4}$
$$-z'/4+\frac{1}{4t\ln(t)}z=\frac{t^3}{\ln(t)}$$
$$-z'+\frac{1}{t\ln(t)}z=4\frac{t^3}{\ln(t)}$$
$$(\frac{z}{\ln t})'=-\frac {4{t^3}}{\ln^2 t}$$
$$\frac{z}{\ln t}=-\int \frac {4{t^3}}{\ln^2 t}dt$$
$$z(t)=- \ln t\int \frac {4{t^3}}{\ln^2 t}dt$$
$$\frac 1 {x^4(t)}=- \ln t\int \frac {4{t^3}}{\ln^2 t}dt$$
You can't integrate the last integral with elementary functions

