integral of (2x+1)/(x+1)

Lösung

\int \frac{2 x + 1}{x + 1} d x

2 x - ln ∣ x + 1 ∣ + C

Schritte zur Lösung

\int \frac{2 x + 1}{x + 1} d x

Lange Division

\frac{2 x + 1}{x + 1} : 2 - \frac{1}{x + 1}

= \int 2 - \frac{1}{x + 1} 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 2 d x - \int \frac{1}{x + 1} d x

\int 2 d x = 2 x

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

= 2 x - ln ∣ x + 1 ∣

Füge eine Konstante zur Lösung hinzu

= 2 x - ln ∣ x + 1 ∣ + C

t an g e n t f (x) = \frac{3}{x ^{2}}

\frac{\partial}{\partial x} (\frac{( a + c x )}{c + b})

\int cos (θ) sin (θ) d θ

\frac{d y}{d x} = 5 x + y - 5

\frac{\partial}{\partial u} (u sin (v))