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

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

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integral from 3 to 4 of x/((x-1)^2(x-2))

\int_{3}^{4} \frac{x}{( x - 1 ) ^{2} ( x - 2 )} d x

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

Decimal Notation

0.40869 \dots

Solution steps

\int_{3}^{4} \frac{x}{( x - 1 ) ^{2} ( x - 2 )} d x

Apply u-substitution

: \int_{2}^{3} \frac{u + 1}{u ^{2} ( u - 1 )} d u

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

Take the partial fraction of

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

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

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

\int_{2}^{3} \frac{1}{u ^{2}} d u = \frac{1}{6}

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

= - 2 (ln (3) - ln (2)) - \frac{1}{6} + 2 ln (2)

Simplify

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

= 4 ln (2) - \frac{1}{6} - 2 ln (3)

\int_{3}^{4} \frac{1}{x} d x

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\int_{- 1}^{0} x^{2} - 2 x d x

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