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

解

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

2 ln ∣ t - 1 ∣ + 2 arctan (t) + C

解答ステップ

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

拡張

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

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

\int \frac{2 t ^{2}}{t ^{3} - t ^{2} + t - 1} d t = ln ∣ t - 1 ∣ + \frac{1}{2} ln t^{2} + 1 + arctan (t)

\int \frac{2 t}{t ^{3} - t ^{2} + t - 1} d t = ln ∣ t - 1 ∣ - \frac{1}{2} ln t^{2} + 1 + arctan (t)

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

簡素化

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

= 2 ln ∣ t - 1 ∣ + 2 arctan (t)

定数を解答に追加する

= 2 ln ∣ t - 1 ∣ + 2 arctan (t) + C