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Evaluate: $$\int_{0}^{\infty}\frac{4x\ln (x)}{x^4+4x^2+1}dx$$

I took $x^2$ common from the denominator and then substituted $\ln (x) =u$, and then I was stuck. The result turns out to be $$\int_{-\infty}^{\infty}\frac{4u}{(e^u+e^{-u})^2+2}dx$$

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This integral cannot be calculated with "conventional" methods. The trick I will tell you, is to sub $x=1/t$ and then $dx=-1/t^2dt$. I leave it op to you to do the algebra. You will end up with the negative form of the given integral. In other words, if the given integral is $I$, you end up with $I=-I$ and so $I=0$. This is just the overview of the method. You need to work out the rudiments. That's your exercise otherwise you haven't learned anything from it

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  • $\begingroup$ how did you figure out,that we need to use the above substitution?Like is it applicable on some format of questions or is it just a clicking base substitution? $\endgroup$
    – Atharv
    Jul 10 '20 at 0:32
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    $\begingroup$ The truth is, I have seen problems like this one before. The thing is, you are integrating from zero to infinity, $ln(-x)=-lnx$ and the headcoefficient and constant in the polynomial are the same, being $1$. In short, that is why it works. $\endgroup$
    – imranfat
    Jul 10 '20 at 0:34
  • $\begingroup$ thank you!Can you send some more questions of these types with a link or pdf please.Much appriciated $\endgroup$
    – Atharv
    Jul 10 '20 at 0:36
  • $\begingroup$ These problems do not appear that frequent. They are not "staple" problems. Usually you will find them in the section "improper integrals" in the integration chapter. Any type of $lnx$ in the numerator with a quadratic denominator (with a negative discriminant, why?) where the head coefficient and constant is the same, will work. Integrating from 0 to infinity. Now you can try this out yourself $\endgroup$
    – imranfat
    Jul 10 '20 at 2:43
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Let's write $$\int_0^\infty \frac{x \ln(x)}{x^4+4x^2+1}dx=\int_0^1\frac{x \ln(x)}{x^4+4x^2+1}dx+\int_1^\infty \frac{x \ln(x)}{x^4+4x^2+1}dx$$

For the first integral we change the variable $x\to z=1/x$ and by a simple calculation we show that it is equal to the second integral with a different sign. This approach makes it straightforward that for any $s>1$ we have $$\int_{\frac{1}{s}}^s \frac{x \ln(x)}{x^4+4x^2+1}dx=0$$ Original question concerned the case $s \to \infty$.

The change of variable is obvious for those working with elliptic integrals.

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