integral of (5t-2)/(t+1)

Lösung

\int \frac{5 t - 2}{t + 1} d t

5 (t + 1) - 7 ln ∣ t + 1 ∣ + C

Schritte zur Lösung

\int \frac{5 t - 2}{t + 1} d t

Wende U-Substitution an

= \int \frac{5 u - 7}{u} d u

Multipliziere aus

\frac{5 u - 7}{u} : 5 - \frac{7}{u}

= \int 5 - \frac{7}{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{7}{u} d u

\int 5 d u = 5 u

\int \frac{7}{u} d u = 7 ln ∣ u ∣

= 5 u - 7 ln ∣ u ∣

Setze in

u = t + 1

ein

= 5 (t + 1) - 7 ln ∣ t + 1 ∣

Füge eine Konstante zur Lösung hinzu

= 5 (t + 1) - 7 ln ∣ t + 1 ∣ + C

\frac{d}{d x} (\frac{x}{12} + 1)

\int_{1}^{2} 2 x^{3} - 4 d x

\frac{\partial}{\partial x} (\frac{5 y ^{2}}{x})

in v erse l a pl a ce \frac{2 s}{( s ^{2} + 4 ) ^{2}}

\frac{d}{d x} (\frac{x + 1}{x})