integral from 8 to R of 1/(x^2-9x+18)

Lösung

\int_{8}^{R} \frac{1}{x ^{2} - 9 x + 18} d x

- \frac{1}{3} (ln ∣ R - 3 ∣ - ln (5)) + \frac{1}{3} (ln ∣ R - 6 ∣ - ln (2))

Schritte zur Lösung

\int_{8}^{R} \frac{1}{x ^{2} - 9 x + 18} d x

Ermittle den Partialbruch von

\frac{1}{x ^{2} - 9 x + 18} : - \frac{1}{3 ( x - 3 )} + \frac{1}{3 ( x - 6 )}

= \int_{8}^{R} - \frac{1}{3 ( x - 3 )} + \frac{1}{3 ( x - 6 )} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{8}^{R} \frac{1}{3 ( x - 3 )} d x + \int_{8}^{R} \frac{1}{3 ( x - 6 )} d x

\int_{8}^{R} \frac{1}{3 ( x - 3 )} d x = \frac{1}{3} (ln ∣ R - 3 ∣ - ln (5))

\int_{8}^{R} \frac{1}{3 ( x - 6 )} d x = \frac{1}{3} (ln ∣ R - 6 ∣ - ln (2))

= - \frac{1}{3} (ln ∣ R - 3 ∣ - ln (5)) + \frac{1}{3} (ln ∣ R - 6 ∣ - ln (2))

t an g e n t f (x) = \frac{6}{3 x + 1}, (- 1, - 3)

\int \frac{1}{2} ln (\frac{1 + x}{1 - x}) d x

\int \frac{6 x + 7}{( 9 x ^{2} + 21 x ) ^{4}} d x

n = 1 \sum \infty \frac{1 0 ^{n}}{9 ^{n}}

x \frac{d y}{d x} - y = x^{2} e^{x}