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

解

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

- \frac{1}{2} ln ∣ t ∣ + \frac{1}{3} ln ∣ t - 1 ∣ + \frac{1}{6} ln ∣ t + 2 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{t ^{3} + t ^{2} - 2 t} : - \frac{1}{2 t} + \frac{1}{3 ( t - 1 )} + \frac{1}{6 ( t + 2 )}

= \int - \frac{1}{2 t} + \frac{1}{3 ( t - 1 )} + \frac{1}{6 ( t + 2 )} d t

総和規則を適用する:

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

= - \int \frac{1}{2 t} d t + \int \frac{1}{3 ( t - 1 )} d t + \int \frac{1}{6 ( t + 2 )} d t

\int \frac{1}{2 t} d t = \frac{1}{2} ln ∣ t ∣

\int \frac{1}{3 ( t - 1 )} d t = \frac{1}{3} ln ∣ t - 1 ∣

\int \frac{1}{6 ( t + 2 )} d t = \frac{1}{6} ln ∣ t + 2 ∣

= - \frac{1}{2} ln ∣ t ∣ + \frac{1}{3} ln ∣ t - 1 ∣ + \frac{1}{6} ln ∣ t + 2 ∣

定数を解答に追加する

= - \frac{1}{2} ln ∣ t ∣ + \frac{1}{3} ln ∣ t - 1 ∣ + \frac{1}{6} ln ∣ t + 2 ∣ + C