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

解

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

2 t^{2} - 3 t + 3 arctan (t) + 2 + C

解答ステップ

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

拡張

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

\int \frac{4 t ^{3}}{t ^{2} + 1} d t = 2 (t^{2} + 1 - ln t^{2} + 1)

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

\int \frac{4 t}{t ^{2} + 1} d t = 2 ln t^{2} + 1

= 2 (t^{2} + 1 - ln t^{2} + 1) - 3 (- arctan (t) + t) + 2 ln t^{2} + 1

簡素化

2 (t^{2} + 1 - ln t^{2} + 1) - 3 (- arctan (t) + t) + 2 ln t^{2} + 1 : 2 t^{2} - 3 t + 3 arctan (t) + 2

= 2 t^{2} - 3 t + 3 arctan (t) + 2

定数を解答に追加する

= 2 t^{2} - 3 t + 3 arctan (t) + 2 + C