integral of 8/(x+1)+x

Lösung

\int \frac{8}{x + 1} + x d x

8 ln ∣ x + 1 ∣ + \frac{x ^{2}}{2} + C

Schritte zur Lösung

\int \frac{8}{x + 1} + x 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 \frac{8}{x + 1} d x + \int x d x

\int \frac{8}{x + 1} d x = 8 ln ∣ x + 1 ∣

\int x d x = \frac{x ^{2}}{2}

= 8 ln ∣ x + 1 ∣ + \frac{x ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= 8 ln ∣ x + 1 ∣ + \frac{x ^{2}}{2} + C

\frac{d}{d x} ((3 x + 7)^{2})

\int_{- 9}^{0} y + 9 d y

x \to 3 lim (x^{2} + 8 x - 2)

n = 5 \sum \infty \frac{2}{n ^{2} - 1}

\int \frac{x}{5 x - 2} d x