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Question Number 85601 by sahnaz last updated on 23/Mar/20

∫((4u)/(4u^2 −4u+1))du

$$\int\frac{\mathrm{4u}}{\mathrm{4u}^{\mathrm{2}} −\mathrm{4u}+\mathrm{1}}\mathrm{du} \\ $$

Commented by Tony Lin last updated on 23/Mar/20

∫((4u)/(4u^2 −4u+1))du =(1/2)∫((8u−4)/(4u^2 −4u+1))du+2∫(1/((2u−1)^2 ))du =(1/2)ln(2u−1)^2 −(1/(2u−1))+c

$$\int\frac{\mathrm{4}{u}}{\mathrm{4}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{8}{u}−\mathrm{4}}{\mathrm{4}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}}{du}+\mathrm{2}\int\frac{\mathrm{1}}{\left(\mathrm{2}{u}−\mathrm{1}\right)^{\mathrm{2}} }{du}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}{u}−\mathrm{1}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}{u}−\mathrm{1}}+{c} \\ $$

Commented by mathmax by abdo last updated on 23/Mar/20

I =∫ ((4u)/(4u^2 −4u+1))du =4∫ ((udu)/((2u−1)^2 )) =2 ∫ ((2u−1+1)/((2u−1)^2 ))du =2 ∫(du/(2u−1)) +2 ∫ (du/((2u−1)^2 )) =ln∣2u−1∣ −(1/(2u−1)) +C

$${I}\:=\int\:\frac{\mathrm{4}{u}}{\mathrm{4}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}}{du}\:=\mathrm{4}\int\:\:\frac{{udu}}{\left(\mathrm{2}{u}−\mathrm{1}\right)^{\mathrm{2}} }\:=\mathrm{2}\:\int\:\frac{\mathrm{2}{u}−\mathrm{1}+\mathrm{1}}{\left(\mathrm{2}{u}−\mathrm{1}\right)^{\mathrm{2}} }{du} \\ $$$$=\mathrm{2}\:\int\frac{{du}}{\mathrm{2}{u}−\mathrm{1}}\:+\mathrm{2}\:\int\:\:\frac{{du}}{\left(\mathrm{2}{u}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$={ln}\mid\mathrm{2}{u}−\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{2}{u}−\mathrm{1}}\:+{C} \\ $$

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