integral of log_{2}(x+2)

Lösung

\int lo g_{2} (x + 2) d x

x lo g_{2} (x + 2) - \frac{1}{ln ( 2 )} (x + 2 - 2 ln ∣ x + 2 ∣) + C

Schritte zur Lösung

\int lo g_{2} (x + 2) d x

Wende die partielle Integration an

= x lo g_{2} (x + 2) - \int \frac{x}{ln ( 2 ) ( x + 2 )} d x

\int \frac{x}{ln ( 2 ) ( x + 2 )} d x = \frac{1}{ln ( 2 )} (x + 2 - 2 ln ∣ x + 2 ∣)

= x lo g_{2} (x + 2) - \frac{1}{ln ( 2 )} (x + 2 - 2 ln ∣ x + 2 ∣)

Füge eine Konstante zur Lösung hinzu

= x lo g_{2} (x + 2) - \frac{1}{ln ( 2 )} (x + 2 - 2 ln ∣ x + 2 ∣) + C

\int_{0}^{1} e^{x} - 1 d x

\int sin (ln (6 x)) d x

\frac{d}{d x} (2 \cdot 3 x)

d er i v a t i v e f (x) = - 4 x + 8

n = 5 \sum \infty \frac{7 ^{n}}{8 ^{n}}