How to set up partial fractions? I have set up partial fractions so that$$A\ln^3x-B(x+x^2)=1-x^2$$ and $$ C\ln^3x+D(x+x^2)=1+x^2$$ to set up and solve the following  $$\alpha(1+x)+ \gamma x= A+C$$ and from $$\frac {D \ln x-B}{\ln^3x} \to  $$ $$\frac {\Delta \ln x}{\ln^2 x}-\frac {\beta}{\ln x}\to$$ $$\frac{\Delta}{\ln x}-\frac{\beta}{\ln x}=\frac {D \ln x-B}{\ln^3x}$$
Assuming A,B,C,and, D are either constants or functions of x, and  $\alpha,\beta,\gamma,\space and \space \Delta$ are likewise constants or functions of x, how do I solve for these?
Guess I should post the rest of my work (I didn't because I did not want to make a duplicate.
Note this work comes from  my attempt to solve  Oskansa's Integral
Here it is
Start with the original derivative $$ \frac {(1-x^2)+(1+x^2)\ln x}{(x+x^2)\ln^3x}dx$$ Break the fraction in the normal way so that  you have $$\frac{(1-x^2)}{(x+x^2)\ln^3x}+\frac{(1+x^2)}{(x+x^2)\ln^2x)}$$ Now break it again so that $$\frac{A}{x+x^2}-\frac{B}{\ln^3x}=\frac{(1-x^2)}{(x+x^2)\ln^3x} $$ and $$\frac{C}{x+x^2}+\frac{D}{\ln^2x}=\frac{(1+x^2)}{(x+x^2)\ln^2x)}$$ So now the derivative is $$\frac{A}{x+x^2}-\frac{B}{\ln^3x}+ \frac{C}{x+x^2}+\frac{D}{\ln^2x}dx=$$ $$\frac{A+C}{x+x^2}+ \frac{D\ln x-B}{\ln^3x}dx$$ Then break them again so that $$\frac {\alpha}{x}+\frac{\gamma}{1+x}= \frac{A+C}{x+x^2}$$ and $$\frac {\nu lnx}{ln^2x}-\frac {\beta}{\ln x}= \frac {\nu}{\ln x}-\frac {\beta}{\ln x}=\frac{D\ln x-B}{\ln^3x}dx$$ then the integral becomes $$I =\int \frac {\alpha}{x}+\int\frac{\gamma}{1+x}+\int \frac {\nu}{\ln x}-\int\frac{\beta}{\ln x}\space dx=$$ 
NOTE $\nu$ in this solution =$\Delta$ in this OP.
 A: Note: To use partial fraction decomposition,  the numerator and denominator must be polynomials. (I.e., partial fractions work with rational functions.) The factor of $\ln^3 x$ in the denominator tells us immediately that the denominator is not a polynomial.

Note that in your originally posted function, the denominator factors to $$(x^2 + x)\ln^3(x) = x(x+1)\ln^3 x$$
I can't remember the numerator, off-hand, but you can split it into two integrals (there were two terms in it). 
Edit
With the edit to the question: Note that the first of the "split integral": $$\frac{(1-x^2)}{(x+x^2)ln^3x} = \dfrac{(1-x)(1+x)}{x(1+x)\ln^3 x} = \dfrac{1-x}{x\ln^3 x} = \dfrac{1}{x\ln^3 x} - \dfrac 1{\ln^3 x}$$ In the first term, you can substitute $u = \ln x \implies du = \dfrac 1x \,dx$. That gives you $$\int \dfrac 1{x\ln^3(x)}\,dx = \int \dfrac{\,du}{u^3} = \int u^{-3}\,du$$
A: You can't really use partial fractions for this.  The standard method of partial fractions only works if the original function is a rational function, that is, a polynomial divided by a polynomial.  The $\ln$ term throws this right out.
You could possibly get something like partial fractions when other than polynomials are involved, but as far as I know there is no systematic way of doing it.  You would have to use trial and error, and I should say the chance of success would be pretty small.
However you don't need all this to do the integral - it is actually a fairly routine integral which has been dressed up to make it look harder.  Hint.  Factorise the polynomials in the numerator and denominator and see what you notice.
Note.  I'm referring to the integral in the title of the question but it seems that it has just been removed. . . so I'm no longer entirely sure what you want to ask.
