What is the integral of (w-1)/(2w^2-2w+1) ?

The integral of (w-1)/(2w^2-2w+1) is 1/4 ln|4w^2-4w+2|-1/2 arctan(2w-1)+C

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integral of (w-1)/(2w^2-2w+1)

\int \frac{w - 1}{2 w ^{2} - 2 w + 1} d w

\frac{1}{4} ln 4 w^{2} - 4 w + 2 - \frac{1}{2} arctan (2 w - 1) + C

Solution steps

\int \frac{w - 1}{2 w ^{2} - 2 w + 1} d w

Complete the square

2 w^{2} - 2 w + 1 : 2 (w - \frac{1}{2})^{2} + \frac{1}{2}

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

Apply u-substitution

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

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

Expand

\frac{2 u - 1}{4 u ^{2} + 1} : \frac{2 u}{4 u ^{2} + 1} - \frac{1}{4 u ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= \frac{1}{4} ln 4 u^{2} + 1 - \frac{1}{2} arctan (2 u)

Substitute back

u = w - \frac{1}{2}

= \frac{1}{4} ln 4 (w - \frac{1}{2})^{2} + 1 - \frac{1}{2} arctan (2 (w - \frac{1}{2}))

Simplify

\frac{1}{4} ln 4 (w - \frac{1}{2})^{2} + 1 - \frac{1}{2} arctan (2 (w - \frac{1}{2})) : \frac{1}{4} ln 4 w^{2} - 4 w + 2 - \frac{1}{2} arctan (2 w - 1)

= \frac{1}{4} ln 4 w^{2} - 4 w + 2 - \frac{1}{2} arctan (2 w - 1)

Add a constant to the solution

= \frac{1}{4} ln 4 w^{2} - 4 w + 2 - \frac{1}{2} arctan (2 w - 1) + C

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What is the integral of (w-1)/(2w^2-2w+1) ?
The integral of (w-1)/(2w^2-2w+1) is 1/4 ln|4w^2-4w+2|-1/2 arctan(2w-1)+C