4u-4u-2-4u-1-du-

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

\int \frac{4 u}{{4 u}^{2} - 4 u + 1} du

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

\int \frac{4 u}{4 u^{2} - 4 u + 1} d u

= \frac{1}{2} \int \frac{8 u - 4}{4 u^{2} - 4 u + 1} d u + 2 \int \frac{1}{{(2 u - 1)}^{2}} d u

= \frac{1}{2} l n {(2 u - 1)}^{2} - \frac{1}{2 u - 1} + c

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

I = \int \frac{4 u}{4 u^{2} - 4 u + 1} d u = 4 \int \frac{u d u}{{(2 u - 1)}^{2}} = 2 \int \frac{2 u - 1 + 1}{{(2 u - 1)}^{2}} d u

= 2 \int \frac{d u}{2 u - 1} + 2 \int \frac{d u}{{(2 u - 1)}^{2}}

= l n ∣ 2 u - 1 ∣ - \frac{1}{2 u - 1} + C