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

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

Commented by MJS_new last updated on 02/Dec/20

$$\frac{4x^2}{x^4+1}=\frac{\frac{\sqrt{2}}{2}(1-\mathrm{i})}{x-\frac{\sqrt{2}}{2}(1+\mathrm{i})}+\frac{\frac{\sqrt{2}}{2}(1+\mathrm{i})}{x-\frac{\sqrt{2}}{2}(1-\mathrm{i})}-\frac{\frac{\sqrt{2}}{2}(1+\mathrm{i})}{x+\frac{\sqrt{2}}{2}(1-\mathrm{i})}-\frac{\frac{\sqrt{2}}{2}(1-\mathrm{i})}{x+\frac{\sqrt{2}}{2}(1+\mathrm{i})}$$$$\Rightarrow \mathrm{we}\mathrm{have}\mathrm{to}\mathrm{solve}$$$$a\int \frac{\ln x}{x-b}dx$$$$\mathrm{and}\mathrm{then}\mathrm{insert}$$$$\mathrm{I}\mathrm{get}a\int \frac{\ln x}{x-b}dx=a(\ln (-b)\ln \mid x+b\mid -{\mathrm{Li}}_2(\frac{x}{b}+1))+C$$$$\mathrm{but}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}\mathrm{if}\mathrm{this}\mathrm{is}\mathrm{right}\mathrm{for}a,b\notin \mathbb{R}$$

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

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

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