What is the integral from 2 to 4 of 1/(x^2-6x+5) ?

The integral from 2 to 4 of 1/(x^2-6x+5) is -1/2 ln(3)

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integral from 2 to 4 of 1/(x^2-6x+5)

\int_{2}^{4} \frac{1}{x ^{2} - 6 x + 5} d x

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

Decimal

- 0.54930 \dots

Solution steps

\int_{2}^{4} \frac{1}{x ^{2} - 6 x + 5} d x

Take the partial fraction of

\frac{1}{x ^{2} - 6 x + 5} : - \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x - 5 )}

= \int_{2}^{4} - \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x - 5 )} 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}^{4} \frac{1}{4 ( x - 1 )} d x + \int_{2}^{4} \frac{1}{4 ( x - 5 )} d x

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

\int_{2}^{4} \frac{1}{4 ( x - 5 )} d x = - \frac{1}{4} ln (3)

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

Simplify

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

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

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What is the integral from 2 to 4 of 1/(x^2-6x+5) ?
The integral from 2 to 4 of 1/(x^2-6x+5) is -1/2 ln(3)