What is the integral from 0 to 1 of (x-6)/(x^2-6x+8) ?

The integral from 0 to 1 of (x-6)/(x^2-6x+8) is -ln(3)

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integral from 0 to 1 of (x-6)/(x^2-6x+8)

\int_{0}^{1} \frac{x - 6}{x ^{2} - 6 x + 8} d x

- ln (3)

Decimal

- 1.09861 \dots

Solution steps

\int_{0}^{1} \frac{x - 6}{x ^{2} - 6 x + 8} d x

Take the partial fraction of

\frac{x - 6}{x ^{2} - 6 x + 8} : \frac{2}{x - 2} - \frac{1}{x - 4}

= \int_{0}^{1} \frac{2}{x - 2} - \frac{1}{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_{0}^{1} \frac{2}{x - 2} d x - \int_{0}^{1} \frac{1}{x - 4} d x

\int_{0}^{1} \frac{2}{x - 2} d x = - 2 ln (2)

\int_{0}^{1} \frac{1}{x - 4} d x = ln (3) - 2 ln (2)

= - 2 ln (2) - (ln (3) - 2 ln (2))

Simplify

= - ln (3)

3 x \frac{d y}{d x} = x^{2} - 4

x \to 4 lim (2 x^{2} - 2 x)

\int_{- \infty}^{6} r e^{\frac{r}{2}} d r

\frac{\partial}{\partial x} (4 y^{3})

\frac{d y}{d t} = 3 y + e^{7 t}

What is the integral from 0 to 1 of (x-6)/(x^2-6x+8) ?
The integral from 0 to 1 of (x-6)/(x^2-6x+8) is -ln(3)