Find the value of the following integral $$\int\frac{4x\ln(x)}{3x^2+2x+1}dx$$ I have no idea how to solve this integral. I tried all the ways of doing it(parts, substitution and many more) but I got stuck every time at the beginning. Please help me to solve this. By the way, calculations involving complex numbers, polylogarithms and special integrals are allowed.
Any help will be greatly appreciated.
EDIT: There are no bounds on the integral. EDIT: I'm asking for the anti derivative.
On
$$ \int\frac{4x\ln x}{3x^{2}+2x+1} = \ln x\int\frac{4x}{3x^{2}+2x+1}dx - \int\frac{1}{x}\int\frac{4x}{3x^{2}+2x+1}dxdx. $$
This is by integration by parts.
Recall $$ \int \frac{dx}{x^{2}+1} = \arctan(x) + C $$
so you can complete the square of the denominator: $$ 3x^{2} + 2x + 1 = (\sqrt{3}x+\frac{1}{\sqrt{3}})^{2} + \frac{2}{3}. $$ Therefore $$ \int\frac{1}{3x^{2}+2x+1}dx = \int\frac{dx}{(\sqrt{3}x+\frac{1}{\sqrt{3}})^{2} + \frac{2}{3}} = \int\frac{dx}{3(x+\frac{1}{3})^{2} + \frac{2}{3}} = (u = x - \frac{1}{3}) = \int\frac{du}{3u^{2}+\frac{2}{3}} = \frac{1}{3}\int\frac{du}{u^{2}+\frac{2}{9}} = \frac{3}{2}\int\frac{du}{\frac{9}{2}u^{2}+1} = (z = \sqrt{\frac{9}{2}}u) =\frac{3}{2}\sqrt{\frac{2}{9}}\int\frac{dz}{z^{2}+1} = \frac{3}{2}\sqrt{\frac{2}{9}}\arctan(z)+ C = \frac{3}{2}\sqrt{\frac{2}{9}}\arctan(\sqrt{\frac{9}{2}}(x-\frac{1}{3}))+ C. $$
Now you can use this knowledge for the first integral. $$\int\frac{x}{3x^{2}+2x+1}dx = x\frac{3}{2}\sqrt{\frac{2}{9}}\arctan(\sqrt{\frac{9}{2}}(x-\frac{1}{3})) - \int\frac{3}{2}\sqrt{\frac{2}{9}}\arctan(\sqrt{\frac{9}{2}}(x-\frac{1}{3}))dx.$$
It is well known that $$ \int \arctan(x)dx = x\arctan(x) - \frac{1}{2}\ln|1+x^{2}| + C. $$
By similar tricks to those I used to derive the integral you may solve the second one. Do not forget to multiply both results by $4$ in the end.
I think you've just gotta brute force it. Since $$ 3x^2 +2x+1 = 3(x- \alpha)(x-\overline{\alpha}) $$ where $\alpha = -\frac{1}{3} + \frac{\sqrt{2}}{3}i$ is one of the roots of the polynomial, you can apply partial fractions to get $$ \frac{x}{(x-\alpha)(x-\overline{\alpha})} = \frac{\alpha}{\alpha - \overline{\alpha}} \frac{1}{x-\alpha}+\frac{\overline{\alpha}}{ \overline{\alpha}-\alpha} \frac{1}{x-\overline{\alpha}} $$ Then the problem reduces to evaluating $\int \frac{\ln(x)}{x-a}\mathrm{d}x$. For this last family of antiderivatives frist recall that $$ \operatorname{Li}_2(t) = -\int\frac{\ln(1-t)}{t}\, \mathrm{d}t $$ So we get \begin{align} \int \frac{\ln(x)}{x-a}\, \mathrm{d}x& = \frac{1}{a}\int \frac{\ln(x)}{\frac{x}{a}-1}\, \mathrm{d}x\\ & \overset{u =x/a}{=} \int \frac{\ln(u) + \ln(a)}{u-1}\, \mathrm{d}u \\ & \overset{s = 1-u}{=} \ln(a) \int \frac{1}{u-1}\, \mathrm{d}u + \int \frac{\ln(1-s)}{s}\, \mathrm{d}s \\ & = \ln(a) \ln(u-1) - \operatorname{Li}_2(s)\\ & = \ln(a)\ln\left(\frac{x}{a}-1\right) -\operatorname{Li}_2\left(1- \frac{x}{a}\right) \end{align} And combining everything we get \begin{align} \int \frac{4x\ln(x)}{3x^2 +2x+1}\, \mathrm{d}x &= \frac{4}{3}\int\frac{x\ln(x)}{(x-\alpha)(x-\overline{\alpha})}\, \mathrm{d}x\\ & = \frac{4}{3} \frac{\alpha}{\alpha - \overline{\alpha}} \int \frac{\ln(x)}{x-\alpha} \, \mathrm{d}x +\frac{4}{3} \frac{\overline{\alpha}}{ \overline{\alpha}-\alpha} \int \frac{\ln(x)}{x-\overline{\alpha}} \, \mathrm{d}x\\ & = \frac{4}{3} \frac{\alpha}{\alpha - \overline{\alpha}} \left[ \ln(\alpha)\ln\left(\frac{x}{\alpha}-1\right) -\operatorname{Li}_2\left(1- \frac{x}{\alpha}\right)\right]+\frac{4}{3} \frac{\overline{\alpha}}{ \overline{\alpha}-\alpha}\left[\ln(\overline{\alpha})\ln\left(\frac{x}{\overline{\alpha}}-1\right) -\operatorname{Li}_2\left(1- \frac{x}{\overline{\alpha}}\right) \right] \end{align} where $\alpha = -\frac{1}{3} + \frac{\sqrt{2}}{3}i$ and $\overline{\alpha} = -\frac{1}{3} - \frac{\sqrt{2}}{3}i$.