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

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

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

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

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

Decimal Notation

- 0.27465 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (ln (1 - 6 x))

- x^{2} + 3 x + 4, at x = 1

\frac{d y}{d x} = \frac{- e ^{x}}{8}

\frac{\partial}{\partial y} (x y e^{z})

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