What is the integral from 0 to 1 of x/((x-3)^2) ?

The integral from 0 to 1 of x/((x-3)^2) is ln(2)-ln(3)+1/2

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integral from 0 to 1 of x/((x-3)^2)

\int_{0}^{1} \frac{x}{( x - 3 ) ^{2}} d x

ln (2) - ln (3) + \frac{1}{2}

Decimal Notation

0.09453 \dots

Solution steps

\int_{0}^{1} \frac{x}{( x - 3 ) ^{2}} d x

Apply u-substitution

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

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

Expand

\frac{u + 3}{u ^{2}} : \frac{1}{u} + \frac{3}{u ^{2}}

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

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

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

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

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

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What is the integral from 0 to 1 of x/((x-3)^2) ?
The integral from 0 to 1 of x/((x-3)^2) is ln(2)-ln(3)+1/2