<s>[INST] Integral $\int_{a}^{b}\frac{x\ln(x)}{x^4+a^{2}b^{2}}dx$ 
How to evaluate the following integral $$\int_{a}^{b}\frac{x\ln(x)}{x^4+a^{2}b^{2}}dx?$$ where $0<a<b$

I observed that we can change the variable in a way that simplifies the expression but the coefficients of the integral becomes extremely ugly. 
With the substitution $t=x^2$ it ends being $$\frac{1}{4}\int_{a^2}^{b^2} \frac{\ln t}{t^{2}+a^{2}b^{2}}dt=\frac14\int_{a^2}^{b^2}\frac{\ln t}{(t-ab)^2-2abt}dt$$ 
 [/INST] $$I=\int_ {a}^{b}\frac{x\ln(x)}{x^4+a^{2}b^{2}}dx\overset{x^2=t}=\frac14 \int_{a^2}^{b^2} \frac{\ln t}{t^2+a^2b^2}dt $$
Let's get rid of that ugly denominator by doing a $t=y(ab)$:
$$I=\frac{ab}4\int_\frac{a}{b}^\frac{b}{a}\frac{\ln(yab)}{a^2b^2(y^2+1) }dy=\frac{1}{4ab}\int_\frac{a}{b}^\frac{b}{a}\frac{\color{blue}{\ln y+\ln(ab)}}{y^2+1 }dy$$
And now one might notice the simmetry directly. Substituting $y=\frac{1}{x}$ gives:
$$\frac{1}{4ab}\int_\frac{a}{b}^\frac{b}{a}\frac{\color{red}{\ln\left(\frac{1}{x}\right)+\ln(ab)}}{x^2+1 }dx$$
Adding the red and blue integral simiplifies quite nice the logarithm:
$$\require{cancel} 2I=\frac{1}{4ab}\int_\frac{a}{b}^\frac{b}{a}\frac{\color{red}{\cancel{\ln\left(\frac{1}{x}\right)}+\ln(ab)}+\color{blue}{\cancel{\ln x+}\ln(ab)}}{x^2+1}dx$$
$$\Rightarrow I=\frac{\ln (ab)}{4ab}\int_\frac{a}{b}^\frac{b}{a}\frac{1}{x^2+1}dx=\frac{\ln (ab)}{4ab}\left(\arctan\frac{b}{a} -\arctan \frac{a}{b}\right)$$
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