integral of (5t-3)/(t+1)

Lösung

\int \frac{5 t - 3}{t + 1} d t

5 (t + 1) - 8 ln ∣ t + 1 ∣ + C

Schritte zur Lösung

\int \frac{5 t - 3}{t + 1} d t

Wende U-Substitution an

= \int \frac{5 u - 8}{u} d u

Multipliziere aus

\frac{5 u - 8}{u} : 5 - \frac{8}{u}

= \int 5 - \frac{8}{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 5 d u - \int \frac{8}{u} d u

\int 5 d u = 5 u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

= 5 u - 8 ln ∣ u ∣

Setze in

u = t + 1

ein

= 5 (t + 1) - 8 ln ∣ t + 1 ∣

Füge eine Konstante zur Lösung hinzu

= 5 (t + 1) - 8 ln ∣ t + 1 ∣ + C

n = 0 \sum \infty \frac{1}{( 2 n + 1 ) !}

\frac{\partial}{\partial x} (2 x e^{x y})

x \to 6 lim (\frac{3 x}{x ^{2} - 6 x})

\int_{1}^{\infty} x d x

\frac{d y}{d x} = \frac{y}{x} + 3 x + 7