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$$\int \frac{x^2(1-\ln(x))}{\ln^4(x)-x^4} dx$$ I tried to factorize the denominator so that I could apply integration by parts but that didn't help me at all.

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    $\begingroup$ Could we stop fighting in here $\endgroup$
    – Tejus
    Sep 18, 2016 at 12:19

2 Answers 2

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HINT:

$$\mathcal{I}(x)=\int\frac{x^2\left(1-\ln(x)\right)}{\ln^4(x)-x^4}\space\text{d}x=$$ $$\frac{1}{2}\int\frac{\ln(x)-1}{x^2+\ln^2(x)}\space\text{d}x-\frac{1}{4}\int\frac{1+x}{x\left(x+\ln(x)\right)}\space\text{d}x+\frac{1}{4}\int\frac{1-x}{x\left(\ln(x)-x\right)}\space\text{d}x$$

  1. Now, for $\int\frac{1-x}{x\left(\ln(x)-x\right)}\space\text{d}x$, substitute $u=\ln(x)-x$ and $\text{d}u=\left(\frac{1}{x}-1\right)\space\text{d}x$: $$\int\frac{1-x}{x\left(\ln(x)-x\right)}\space\text{d}x=\int\frac{1}{u}\space\text{d}u=\ln\left|u\right|+\text{C}=\ln\left|\ln(x)-x\right|+\text{C}$$
  2. Now, for $\int\frac{1+x}{x\left(x+\ln(x)\right)}\space\text{d}x$, substitute $s=\ln(x)+x$ and $\text{d}s=\left(\frac{1}{x}-1\right)\space\text{d}x$: $$\int\frac{1+x}{x\left(x+\ln(x)\right)}\space\text{d}x=\int\frac{1}{s}\space\text{d}s=\ln\left|s\right|+\text{C}=\ln\left|\ln(x)+x\right|+\text{C}$$
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    $\begingroup$ THAT is an answer!! Great as always :) $\endgroup$
    – Laplacian
    Sep 18, 2016 at 12:23
  • $\begingroup$ @FourierTransform Thx :) $\endgroup$ Sep 18, 2016 at 12:24
  • $\begingroup$ What about the third integral ?How do I solve that ? $\endgroup$
    – Tejus
    Sep 18, 2016 at 12:27
  • $\begingroup$ @Tejus Use you integration skills :) and think about it $\endgroup$ Sep 18, 2016 at 12:28
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$$I=\int \frac{x^2(1-\ln(x))}{\ln^4(x)-x^4} dx$$

$$I=\int\frac{\frac{1-\ln x}{x^2}}{\left(\frac{\ln x }{x}\right)^4-1}dx$$

Now put $\displaystyle \frac{\ln x}{x}=t,$ Then $\displaystyle \frac{1-\ln x}{x^2}dx=dt$

So $$$I=\int\frac{1}{t^4-1}dt = \int\frac{1}{(1-t^2)(1+t^2)}dt =\frac{1}{2}\int\left[\frac{1}{1-t^2}+\frac{1}{1+t^2}\right]dt$$

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  • $\begingroup$ This is a beautiful solution . $\endgroup$
    – Tejus
    Sep 18, 2016 at 12:46
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    $\begingroup$ Couldn't be a better answer to a good question like this. I up voted both the post and the answer. $\endgroup$
    – Zlatan
    Sep 19, 2016 at 0:19

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