Question Number 157420 by Gbenga last updated on 23/Oct/21
$$\int\frac{\mathrm{1}}{\boldsymbol{{u}}^{\mathrm{4}} +\left(\mathrm{1}−\boldsymbol{{u}}\right)^{\mathrm{4}} }\boldsymbol{{du}} \\ $$
Answered by EvaNelle00 last updated on 23/Oct/21
$$\int\:\frac{{du}}{{u}^{\mathrm{4}} \left(\mathrm{1}\:+\:\left(\frac{\mathrm{1}−{u}}{{u}}\right)^{\mathrm{4}} \right)} \\ $$$$\int\:\frac{{du}}{{u}^{\mathrm{4}} \left(\mathrm{1}\:+\:\left(\frac{\mathrm{1}}{{u}}−\mathrm{1}\right)^{\mathrm{4}} \right)} \\ $$$$\int\:\frac{{du}}{{u}^{\mathrm{4}} \left(\mathrm{1}\:+\:\left(\mathrm{1}−\frac{\mathrm{1}}{{u}}\right)^{\mathrm{4}} \right)} \\ $$$$\int\left(−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\frac{{du}}{\left(\mathrm{1}\:+\:\left(\mathrm{1}−\frac{\mathrm{1}}{{u}}\right)^{\mathrm{4}} \right)} \\ $$$${x}=\frac{\mathrm{1}}{{u}}\:\Rightarrow{dx}\:=\:−\frac{{du}}{{u}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow−{u}^{\mathrm{2}} {dx}\:=\:{du} \\ $$$$−\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\:\frac{−\frac{{du}}{{u}^{\mathrm{2}} }}{\left(\mathrm{1}\:+\:\left(\mathrm{1}−{x}\right)^{\mathrm{4}} \right)} \\ $$$$−\int\:\frac{{x}^{\mathrm{2}} {dx}}{\left(\mathrm{1}\:+\:\left(\mathrm{1}−{x}\right)^{\mathrm{4}} \right)} \\ $$$$ \\ $$$$\int \\ $$
Answered by MJS_new last updated on 23/Oct/21
$$\int\frac{{du}}{{u}^{\mathrm{4}} +\left(\mathrm{1}−{u}\right)^{\mathrm{4}} }= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\right){du} \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{formula} \\ $$
Commented by Gbenga last updated on 23/Oct/21
$$\boldsymbol{\mathrm{thanks}} \\ $$