What is the integral from 1 to 2 of (x-1)/((2x-1)^2) ?

The integral from 1 to 2 of (x-1)/((2x-1)^2) is -1/6+1/4 ln(3)

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integral from 1 to 2 of (x-1)/((2x-1)^2)

\int_{1}^{2} \frac{x - 1}{( 2 x - 1 ) ^{2}} d x

- \frac{1}{6} + \frac{1}{4} ln (3)

Decimal

0.10798 \dots

Solution steps

\int_{1}^{2} \frac{x - 1}{( 2 x - 1 ) ^{2}} d x

Expand

\frac{x - 1}{( 2 x - 1 ) ^{2}} : \frac{x}{( 2 x - 1 ) ^{2}} - \frac{1}{( 2 x - 1 ) ^{2}}

Apply the Sum Rule:

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

= \int_{1}^{2} \frac{x}{( 2 x - 1 ) ^{2}} d x - \int_{1}^{2} \frac{1}{( 2 x - 1 ) ^{2}} d x

\int_{1}^{2} \frac{x}{( 2 x - 1 ) ^{2}} d x = \frac{1}{4} ln (3) + \frac{1}{6}

\int_{1}^{2} \frac{1}{( 2 x - 1 ) ^{2}} d x = \frac{1}{3}

= \frac{1}{4} ln (3) + \frac{1}{6} - \frac{1}{3}

Combine the fractions

\frac{1}{6} - \frac{1}{3} : - \frac{1}{6}

= - \frac{1}{6} + \frac{1}{4} ln (3)

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What is the integral from 1 to 2 of (x-1)/((2x-1)^2) ?
The integral from 1 to 2 of (x-1)/((2x-1)^2) is -1/6+1/4 ln(3)