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

解

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

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

解答ステップ

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

拡張

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

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

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

\int \frac{2 t ^{2}}{t ^{3} - t ^{2}} d t = 2 ln ∣ t - 1 ∣

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

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

簡素化

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

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

定数を解答に追加する

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