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Question Number 157420 by Gbenga last updated on 23/Oct/21

$$\int \frac{1}{{\mathit{u}}^4+{(1-\mathit{u})}^4}\mathit{d}\mathit{u}$$

Answered by EvaNelle00 last updated on 23/Oct/21

$$\int \frac{du}{u^4(1+{\left(\frac{1-u}{u}\right)}^4)}$$$$\int \frac{du}{u^4(1+{(\frac{1}{u}-1)}^4)}$$$$\int \frac{du}{u^4(1+{(1-\frac{1}{u})}^4)}$$$$\int {(-\frac{1}{u^2})}^2\frac{du}{(1+{(1-\frac{1}{u})}^4)}$$$$x=\frac{1}{u}\Rightarrow dx=-\frac{du}{u^2}$$$$\Rightarrow -u^{2}dx=du$$$$-\int \frac{1}{u^2}\frac{-\frac{du}{u^2}}{(1+{(1-x)}^4)}$$$$-\int \frac{x^{2}dx}{(1+{(1-x)}^4)}$$$$$$$$\int$$

Answered by MJS_new last updated on 23/Oct/21

$$\int \frac{du}{u^4+{(1-u)}^4}=$$$$=\frac{\sqrt{2}}{4}(\int \frac{1}{u^2-u+1-\frac{\sqrt{2}}{2}}-\frac{1}{u^2-u+1+\frac{\sqrt{2}}{2}})du$$$$\mathrm{now}\mathrm{use}\mathrm{formula}$$

Commented by Gbenga last updated on 23/Oct/21

$$\mathbf{thanks}$$

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