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

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

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

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

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

Decimal

0.44131 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{u + 1}{4 u ^{2}} d u

= \int_{1}^{3} \frac{u + 1}{4 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{1}^{3} \frac{u + 1}{u ^{2}} d u

Expand

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

= \frac{1}{4} \cdot \int_{1}^{3} \frac{1}{u} + \frac{1}{u ^{2}} d u

Apply the Sum Rule:

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

= \frac{1}{4} (\int_{1}^{3} \frac{1}{u} d u + \int_{1}^{3} \frac{1}{u ^{2}} d u)

\int_{1}^{3} \frac{1}{u} d u = ln (3)

\int_{1}^{3} \frac{1}{u ^{2}} d u = \frac{2}{3}

= \frac{1}{4} (ln (3) + \frac{2}{3})

Simplify

\frac{1}{4} (ln (3) + \frac{2}{3}) : \frac{1}{4} ln (3) + \frac{1}{6}

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

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