integral of (2u^2-1)/(u^3-u)

解

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

ln ∣ u ∣ + \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣ + C

解答ステップ

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

以下の部分分数を得る:

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

= ln ∣ u ∣ + \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣

定数を解答に追加する

= ln ∣ u ∣ + \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣ + C