Evaluate $\int \frac{x^2(\ln x-1)}{x^4-\ln^4x} dx $

Question

Evaluate $$\int \frac{x^2(\ln x-1)}{x^4-\ln^4x} dx $$

Now, I've tried various things like : factoring out $x^4$ to try to do a substitution after . Also, wolfram alpha gives an ugly solution, any thoughts on how to reach that ? Any help will be appreciated !

Answer 1

$$I=\int \frac{x^2(\ln x-1)}{x^4-\ln^4x}dx$$ Let $x=e^t \implies dx= e^t dt$, then $$I=\int \frac{e^{3t}(t-1)}{e^{4t}-t^4}dt=\int\frac{{e^{-t}}(t-1)}{1-(te^{-t})^4}dt$$ Let $te^{-t}=u \implies e^{-t}-t e^{-t}$, then $$I=\int \frac{-du}{1-u^4}du=-\frac{1}{2} \int\left( \frac{1}{1-u^2}+\frac{1}{1+u^2} \right)du=0 \implies I=\frac{1}{4} \ln\frac{1-u}{1+u}-\frac{1}{2} \tan^{-1} u+C$$ Finally, $$I=\frac{1}{4}\ln \frac{x-\ln x}{x+ln x}-\frac{1}{2} \tan^{-1}(\frac{\ln x}{x})+C.$$

Answer 2

Mathematica gives:

$$\frac{1}{4} \ln (x-\ln (x))-\frac{1}{4} \ln (x+\ln (x))-\frac{1}{2} \tan ^{-1}\left(\frac{\ln (x)}{x}\right)$$

I would factor the denominator (it is the difference of two squares) and then use partial fractions.

Answer 3

Set $u=\dfrac{\ln(x)}{x}$

Then $du=\dfrac{1-\ln(x)}{x^2}dx$

The integral becomes

$\displaystyle \int \dfrac{x^4(-du)}{x^4-\ln(x)^4}=\int\dfrac{-du}{1-\frac{\ln(x)^4}{x^4}}=-\int \dfrac {du}{1-u^4}=-\dfrac 12\left(\tan^{-1}(u)+\tanh^{-1}(u)\right)$

Answer 4

We can rewrite the integral as \begin{align} I & = \int \frac{x^2 (\ln x - 1)}{x^4 - \ln^4 x} dx \\ & = \frac{1}{2}\int \frac{((x^2 - \ln^2 x) + (x^2 + \ln^2 x)) (\ln x - 1)}{(x^2 - \ln^2 x)(x^2 + \ln^2 x)} dx \\ & = \frac{1}{2}\int \frac{(\ln x - 1)}{x^2 + \ln^2 x} dx + \frac{1}{2}\int \frac{(\ln x - 1)}{x^2 + \ln^2 x} dx \\ & = \frac{1}{2}\int \frac{(\ln x - 1)}{x^2} \frac{1}{1 + \left(\frac{\ln x}{x}\right)^2} dx + \frac{1}{2}\int \frac{(\ln x - 1)}{x^2} \frac{1}{1 - \left(\frac{\ln x}{x}\right)^2} dx \\ \end{align}

We can now substitute $u = \dfrac{\ln x}{x}$ and get $du = \dfrac{1 - \ln x}{x^2} dx$ to get $$ I = -\frac{1}{2}\int \frac{1}{1 + u^2} du - \frac{1}{2}\int \frac{1}{1 - u^2} du $$

Both of these are now standard integrals. I guess it is straight forward from here.