integral from 0 to x of 1/((1-x)(2-x))

Lösung

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

- ln ∣ x - 1 ∣ + ln ∣ x - 2 ∣ - ln (2)

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int_{0}^{x} \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

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

= - ln ∣ x - 1 ∣ + ln ∣ x - 2 ∣ - ln (2)

\int_{- 2}^{1} (2 x^{2} + 7) d x

\int_{0}^{t} y (t - w) e^{- 4 w} d w

\int_{0}^{1} y e^{- y^{2}} d y

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

\int_{1}^{10} 1