integral of (t^2)/(t+1)

Lösung

\int \frac{t ^{2}}{t + 1} d t

\frac{( t + 1 ) ^{2}}{2} - 2 (t + 1) + ln ∣ t + 1 ∣ + C

Schritte zur Lösung

\int \frac{t ^{2}}{t + 1} d t

Wende U-Substitution an

= \int \frac{( u - 1 ) ^{2}}{u} d u

Multipliziere aus

\frac{( u - 1 ) ^{2}}{u} : u - 2 + \frac{1}{u}

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

\int u d u = \frac{u ^{2}}{2}

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + ln ∣ u ∣

Setze in

u = t + 1

ein

= \frac{( t + 1 ) ^{2}}{2} - 2 (t + 1) + ln ∣ t + 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{( t + 1 ) ^{2}}{2} - 2 (t + 1) + ln ∣ t + 1 ∣ + C

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

\frac{d}{d x} (arccos (- 3 x))

y^{^{''}} + 16 = 0

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

\int \frac{x}{4} d x