integral of (x+1)/(x^2-16)

Lösung

\int \frac{x + 1}{x ^{2} - 16} d x

\frac{3}{8} ln ∣ x + 4 ∣ + \frac{5}{8} ln ∣ x - 4 ∣ + C

Schritte zur Lösung

\int \frac{x + 1}{x ^{2} - 16} d x

Ermittle den Partialbruch von

\frac{x + 1}{x ^{2} - 16} : \frac{3}{8 ( x + 4 )} + \frac{5}{8 ( x - 4 )}

= \int \frac{3}{8 ( x + 4 )} + \frac{5}{8 ( x - 4 )} 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{3}{8 ( x + 4 )} d x + \int \frac{5}{8 ( x - 4 )} d x

\int \frac{3}{8 ( x + 4 )} d x = \frac{3}{8} ln ∣ x + 4 ∣

\int \frac{5}{8 ( x - 4 )} d x = \frac{5}{8} ln ∣ x - 4 ∣

= \frac{3}{8} ln ∣ x + 4 ∣ + \frac{5}{8} ln ∣ x - 4 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{3}{8} ln ∣ x + 4 ∣ + \frac{5}{8} ln ∣ x - 4 ∣ + C

n = 0 \sum \infty 4 n^{2} + 4 n + 1

a re a y = 4 x^{2}, y = x^{2} + 2

x \to 0 lim ((1 - 4 x)^{\frac{1}{x}})

x y^{^{'}} + x y = 1 - y, y (1) = 0

\frac{\partial}{\partial x} (x^{2} y^{2} (2 z + 1)^{2})