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Question Number 133972 by liberty last updated on 26/Feb/21

H = ∫ (((2x−1)^7 )/((2x+1)^9 )) dx

$$\mathscr{H}\:=\:\int\:\frac{\left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{7}} }{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{9}} }\:\mathrm{dx}\: \\ $$

Answered by EDWIN88 last updated on 26/Feb/21

 H = ∫ (((2x−1)/(2x+1)))^7 .(dx/((2x+1)^2 ))  change of variable let u = ((2x−1)/(2x+1))   du = (4/((2x+1)^2 )) dx ; (dx/((2x+1)^2 )) = (du/4)  H = ∫ (u^7 /4) du = (u^8 /(32)) + c = (1/(32))(((2x−1)/(2x+1)))^8  + c

$$\:\mathscr{H}\:=\:\int\:\left(\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2x}+\mathrm{1}}\right)^{\mathrm{7}} .\frac{\mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{change}\:\mathrm{of}\:\mathrm{variable}\:\mathrm{let}\:\mathrm{u}\:=\:\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2x}+\mathrm{1}} \\ $$$$\:\mathrm{du}\:=\:\frac{\mathrm{4}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx}\:;\:\frac{\mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{du}}{\mathrm{4}} \\ $$$$\mathscr{H}\:=\:\int\:\frac{\mathrm{u}^{\mathrm{7}} }{\mathrm{4}}\:\mathrm{du}\:=\:\frac{\mathrm{u}^{\mathrm{8}} }{\mathrm{32}}\:+\:\mathrm{c}\:=\:\underline{\frac{\mathrm{1}}{\mathrm{32}}\left(\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2x}+\mathrm{1}}\right)^{\mathrm{8}} \:+\:\mathrm{c}\:} \\ $$$$ \\ $$

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