integral of (4t^2+3t-1)/(t^3-t^2)

解

\int \frac{4 t ^{2} + 3 t - 1}{t ^{3} - t ^{2}} d t

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

解答ステップ

\int \frac{4 t ^{2} + 3 t - 1}{t ^{3} - t ^{2}} d t

拡張

\frac{4 t ^{2} + 3 t - 1}{t ^{3} - t ^{2}} : \frac{4 t ^{2}}{t ^{3} - t ^{2}} + \frac{3 t}{t ^{3} - t ^{2}} - \frac{1}{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{4 t ^{2}}{t ^{3} - t ^{2}} d t + \int \frac{3 t}{t ^{3} - t ^{2}} d t - \int \frac{1}{t ^{3} - t ^{2}} d t

\int \frac{4 t ^{2}}{t ^{3} - t ^{2}} d t = 4 ln ∣ t - 1 ∣

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

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

= 4 ln ∣ t - 1 ∣ + 3 (- ln ∣ t ∣ + ln ∣ t - 1 ∣) - (- ln ∣ t ∣ + \frac{1}{t} + ln ∣ t - 1 ∣)

簡素化

4 ln ∣ t - 1 ∣ + 3 (- ln ∣ t ∣ + ln ∣ t - 1 ∣) - (- ln ∣ t ∣ + \frac{1}{t} + ln ∣ t - 1 ∣) : 6 ln ∣ t - 1 ∣ - 2 ln ∣ t ∣ - \frac{1}{t}

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

定数を解答に追加する

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