integral of (3t-2)/(t+1)

Lösung

\int \frac{3 t - 2}{t + 1} d t

3 (t + 1) - 5 ln ∣ t + 1 ∣ + C

Schritte zur Lösung

\int \frac{3 t - 2}{t + 1} d t

Wende U-Substitution an

= \int \frac{3 u - 5}{u} d u

Multipliziere aus

\frac{3 u - 5}{u} : 3 - \frac{5}{u}

= \int 3 - \frac{5}{u} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 d u - \int \frac{5}{u} d u

\int 3 d u = 3 u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= 3 u - 5 ln ∣ u ∣

Setze in

u = t + 1

ein

= 3 (t + 1) - 5 ln ∣ t + 1 ∣

Füge eine Konstante zur Lösung hinzu

= 3 (t + 1) - 5 ln ∣ t + 1 ∣ + C

\frac{d}{d x} (in (x))

\int \frac{( 3 + x )}{1 - x ^{2}} d x

a re a 2 x^{3} - x^{2}, 0, 1

\frac{d y}{d t} - 2 y = y^{4}

\frac{d}{d x} ((x - 3) (x^{2} + 6))