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

∫((4u)/(4u^2 −4u+1))du

∫4u4u2−4u+1du

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

∫4u4u2−4u+1du=12∫8u−44u2−4u+1du+2∫1(2u−1)2du=12ln(2u−1)2−12u−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=∫4u4u2−4u+1du=4∫udu(2u−1)2=2∫2u−1+1(2u−1)2du=2∫du2u−1+2∫du(2u−1)2=ln∣2u−1∣−12u−1+C

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