Question Number 133972 by liberty last updated on 26/Feb/21
$$\mathscr{H}\:=\:\int\:\frac{\left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{7}} }{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{9}} }\:\mathrm{dx}\: \\ $$
Answered by EDWIN88 last updated on 26/Feb/21
$$\:\mathscr{H}\:=\:\int\:\left(\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2x}+\mathrm{1}}\right)^{\mathrm{7}} .\frac{\mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{change}\:\mathrm{of}\:\mathrm{variable}\:\mathrm{let}\:\mathrm{u}\:=\:\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2x}+\mathrm{1}} \\ $$$$\:\mathrm{du}\:=\:\frac{\mathrm{4}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx}\:;\:\frac{\mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{du}}{\mathrm{4}} \\ $$$$\mathscr{H}\:=\:\int\:\frac{\mathrm{u}^{\mathrm{7}} }{\mathrm{4}}\:\mathrm{du}\:=\:\frac{\mathrm{u}^{\mathrm{8}} }{\mathrm{32}}\:+\:\mathrm{c}\:=\:\underline{\frac{\mathrm{1}}{\mathrm{32}}\left(\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2x}+\mathrm{1}}\right)^{\mathrm{8}} \:+\:\mathrm{c}\:} \\ $$$$ \\ $$