integral of ln(3x+9)

Lösung

\int ln (3 x + 9) d x

(ln (3 (x + 3)) - 1) (x + 3) + C

Schritte zur Lösung

\int ln (3 x + 9) d x

Wende U-Substitution an

= \int ln (u) \frac{1}{3} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int ln (u) d u

Wende die partielle Integration an

= \frac{1}{3} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{3} (u ln (u) - u)

Setze in

u = 3 x + 9

ein

= \frac{1}{3} ((3 x + 9) ln (3 x + 9) - (3 x + 9))

Vereinfache

\frac{1}{3} ((3 x + 9) ln (3 x + 9) - (3 x + 9)) : (ln (3 (x + 3)) - 1) (x + 3)

= (ln (3 (x + 3)) - 1) (x + 3)

Füge eine Konstante zur Lösung hinzu

= (ln (3 (x + 3)) - 1) (x + 3) + C

\int \frac{sin ( x )}{cos ^{2} ( x ) + cos ( x ) - 2} d x

in v erse l a pl a ce \frac{s ^{2} - 3}{s + 1 \cdot s + 2}

x \to 0 lim (x^{5} e^{- x^{4}})

\frac{\partial}{\partial x} (e^{- 8 t} cos (π x))

a re a - x + 5, x^{2}