integral of (7-x)/(x^2-4x+4)

Lösung

\int \frac{7 - x}{x ^{2} - 4 x + 4} d x

- \frac{7}{x - 2} + \frac{x}{x - 2} - ln ∣ x - 2 ∣ + C

Schritte zur Lösung

\int \frac{7 - x}{x ^{2} - 4 x + 4} d x

Multipliziere aus

\frac{7 - x}{x ^{2} - 4 x + 4} : \frac{7}{x ^{2} - 4 x + 4} - \frac{x}{x ^{2} - 4 x + 4}

Wende die Summenregel an:

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

= \int \frac{7}{x ^{2} - 4 x + 4} d x - \int \frac{x}{x ^{2} - 4 x + 4} d x

\int \frac{7}{x ^{2} - 4 x + 4} d x = - \frac{7}{x - 2}

\int \frac{x}{x ^{2} - 4 x + 4} d x = - \frac{x}{x - 2} + ln ∣ x - 2 ∣

= - \frac{7}{x - 2} - (- \frac{x}{x - 2} + ln ∣ x - 2 ∣)

Vereinfache

= - \frac{7}{x - 2} + \frac{x}{x - 2} - ln ∣ x - 2 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{7}{x - 2} + \frac{x}{x - 2} - ln ∣ x - 2 ∣ + C

\frac{d}{d x} (cos (w x))

\frac{d}{d t} (4 \frac{e ^{5 t}}{t ^{2} + 1})

\frac{y ^{^{''}}}{y} = - 3

\int - 25 sin (- 5 t) d t

\frac{d}{d x} (arccos (e^{6 x}))