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.

  • $\begingroup$ Partial fractions is a tool for rational functions. Rational functions are ratios of two polynomials. Polynomials are sums of powers of $x$. $\ln x$ and $\ln^3 x$ are not powers of $x$. $\endgroup$ – vadim123 Mar 14 '14 at 13:20
  • $\begingroup$ No, but shouldn't I be able to do the same thing I do with $x^3$ in the denominator? That is decompose it to $\frac {a}{x} + \frac {b}{x^2}$? $\endgroup$ – Chris Mar 14 '14 at 13:27
  • $\begingroup$ If your denominator is a polynomial, then you may do partial fractions. If your denominator is not a polynomial, then no. $\endgroup$ – vadim123 Mar 14 '14 at 13:28

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).


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 $$\int \dfrac 1{x\ln^3(x)}\,dx = \int \dfrac{\,du}{u^3} = \int u^{-3}\,du$$

  • $\begingroup$ the factor $ln^3X$ = $ln^2X(lnx)$ $\endgroup$ – Chris Mar 14 '14 at 13:22
  • $\begingroup$ Sure, which is equal to $\ln(X)\ln(X)\ln(X)$, but this is NOT a polynomial! $\endgroup$ – Namaste Mar 14 '14 at 13:23
  • $\begingroup$ Hopefully the edit explains a bit more...though it probably changes nothing $\endgroup$ – Chris Mar 14 '14 at 13:37
  • $\begingroup$ 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 $$\int \dfrac 1{x\ln^3(x)}\,dx = \int \dfrac{\,du}{u^3} = \int u^{-3}\,du$$ $\endgroup$ – Namaste Mar 14 '14 at 13:42
  • $\begingroup$ I see then $x=e^u$ $\endgroup$ – Chris Mar 14 '14 at 17:02

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.

  • $\begingroup$ Yeah well at least it LOOKED pretty. And I learned another rule to ad the many things I can't do. $\endgroup$ – Chris Mar 14 '14 at 13:43

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