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

Lösung

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

ln x^{2} - 16 + ln ∣ x ∣ + 2 x + C

Schritte zur Lösung

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

\int \frac{2 x}{x ^{2} - 16} d x = ln x^{2} - 16

\int \frac{1}{x} d x = ln ∣ x ∣

\int 2 d x = 2 x

= ln x^{2} - 16 + ln ∣ x ∣ + 2 x

Füge eine Konstante zur Lösung hinzu

= ln x^{2} - 16 + ln ∣ x ∣ + 2 x + C

x \to 3 lim (\frac{2 f ( x )}{5})

\int \frac{1}{e ^{t}} d t

\frac{\partial}{\partial x} (\frac{c}{x ^{2}})

n = 5 \sum \infty \frac{4}{7 ^{n}}

\int u^{- \frac{1}{2}} d u