integral of (t^3)/(t^2-4t+4)

Lösung

\int \frac{t ^{3}}{t ^{2} - 4 t + 4} d t

\frac{t ^{2}}{2} + 4 t - \frac{12 t - 16}{t - 2} + 12 ln ∣ t - 2 ∣ + C

Schritte zur Lösung

\int \frac{t ^{3}}{t ^{2} - 4 t + 4} d t

Lange Division

\frac{t ^{3}}{t ^{2} - 4 t + 4} : t + 4 + \frac{12 t - 16}{t ^{2} - 4 t + 4}

= \int t + 4 + \frac{12 t - 16}{t ^{2} - 4 t + 4} d t

Wende die Summenregel an:

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

= \int t d t + \int 4 d t + \int \frac{12 t - 16}{t ^{2} - 4 t + 4} d t

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

\int 4 d t = 4 t

\int \frac{12 t - 16}{t ^{2} - 4 t + 4} d t = - \frac{12 t - 16}{t - 2} + 12 ln ∣ t - 2 ∣

= \frac{t ^{2}}{2} + 4 t - \frac{12 t - 16}{t - 2} + 12 ln ∣ t - 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{t ^{2}}{2} + 4 t - \frac{12 t - 16}{t - 2} + 12 ln ∣ t - 2 ∣ + C

\int 12 x ln (8 x) d x

\int 9 - (x - 1)^{2} d x

\int \frac{2 x - 3}{x ^{2} + 4 x + 20} d x

\int \frac{3}{x 16 - x ^{2}} d x

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