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

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

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

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

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

Decimal

0.12101 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

Simplify

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

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