# Definite integral of the following question [closed]

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 $$\int_{-\infty}^{\infty}\frac{4u}{(e^u+e^{-u})^2+2}dx$$

• Please read how to ask a good question and try editing your post. Without any improvement, others may not be interested in answering, and the question may get closed. Jul 10 '20 at 1:44

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
• 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. Jul 10 '20 at 0:34
• 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 Jul 10 '20 at 2:43
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 $$s \to \infty$$.