I-4x-2-lnx-1-x-4-dx-

Question Number 124277 by SOMEDAVONG last updated on 02/Dec/20

I . \int \frac{{4 x}^{2} lnx}{1 + x^{4}} dx = ? ?

Commented by MJS_new last updated on 02/Dec/20

\frac{4 x^{2}}{x^{4} + 1} = \frac{\frac{\sqrt{2}}{2} (1 - i)}{x - \frac{\sqrt{2}}{2} (1 + i)} + \frac{\frac{\sqrt{2}}{2} (1 + i)}{x - \frac{\sqrt{2}}{2} (1 - i)} - \frac{\frac{\sqrt{2}}{2} (1 + i)}{x + \frac{\sqrt{2}}{2} (1 - i)} - \frac{\frac{\sqrt{2}}{2} (1 - i)}{x + \frac{\sqrt{2}}{2} (1 + i)}

\Rightarrow we have to solve

a \int \frac{\ln x}{x - b} d x

and then insert

I get a \int \frac{\ln x}{x - b} d x = a (\ln (- b) \ln ∣ x + b ∣ - {Li}_{2} (\frac{x}{b} + 1)) + C

but I^{'} m not sure if this is right for a, b \notin R

Commented by Dwaipayan Shikari last updated on 02/Dec/20

\frac{4 x^{2}}{(1 + x^{4})} = \frac{2}{(x^{2} - i)} + \frac{2}{x^{2} + i} = \frac{1}{\sqrt{i}} (\frac{1}{(x - \sqrt{i})} - \frac{1}{(x + \sqrt{i})}) + \frac{1}{i \sqrt{i}} (\frac{1}{x - i \sqrt{i}} - \frac{1}{x + i \sqrt{i}})