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

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

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

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

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

Decimal

0.03606 \dots

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{4} - x ^{2}} : - \frac{1}{x ^{2}} - \frac{1}{2 ( x + 1 )} + \frac{1}{2 ( x - 1 )}

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

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

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

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

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

Simplify

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

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

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