integral of ((u-1))/((u+1))

Lösung

\int \frac{( u - 1 )}{( u + 1 )} d u

- 2 ln ∣ u + 1 ∣ + u + C

Schritte zur Lösung

\int \frac{u - 1}{u + 1} d u

\frac{u - 1}{u + 1} = - \frac{2}{u + 1} + 1

= \int - \frac{2}{u + 1} + 1 d u

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{2}{u + 1} d u + \int 1 d u

\int \frac{2}{u + 1} d u = 2 ln ∣ u + 1 ∣

\int 1 d u = u

= - 2 ln ∣ u + 1 ∣ + u

Füge eine Konstante zur Lösung hinzu

= - 2 ln ∣ u + 1 ∣ + u + C

n = 1 \sum \infty 6 \frac{1}{n}

t an g e n t f (x) = x^{2} + 3 x + 4 π, at x = - 3

\frac{d}{d x} (arccos (\frac{x}{b}))

\frac{d ^{4}}{d x ^{4}} (\frac{8 x}{1 - x})

d er i v a t i v e f (x) = 5^{5 x + 1} + 3^{1 - 3 x}