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

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

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integral from 2 to 3 of 3/(x^2-5x+4)

\int_{2}^{3} \frac{3}{x ^{2} - 5 x + 4} d x

- 2 ln (2)

Decimal Notation

- 1.38629 \dots

Solution steps

\int_{2}^{3} \frac{3}{x ^{2} - 5 x + 4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{2}^{3} \frac{1}{x ^{2} - 5 x + 4} d x

Take the partial fraction of

\frac{1}{x ^{2} - 5 x + 4} : - \frac{1}{3 ( x - 1 )} + \frac{1}{3 ( x - 4 )}

= 3 \cdot \int_{2}^{3} - \frac{1}{3 ( x - 1 )} + \frac{1}{3 ( x - 4 )} 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

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

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

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

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

Simplify

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

= - 2 ln (2)

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