integral of (8t^3-3t^2+8t)/(t^2+1)

解

\int \frac{8 t ^{3} - 3 t ^{2} + 8 t}{t ^{2} + 1} d t

4 t^{2} - 3 t + 3 arctan (t) + 4 + C

解答ステップ

\int \frac{8 t ^{3} - 3 t ^{2} + 8 t}{t ^{2} + 1} d t

拡張

\frac{8 t ^{3} - 3 t ^{2} + 8 t}{t ^{2} + 1} : \frac{8 t ^{3}}{t ^{2} + 1} - \frac{3 t ^{2}}{t ^{2} + 1} + \frac{8 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{8 t ^{3}}{t ^{2} + 1} d t - \int \frac{3 t ^{2}}{t ^{2} + 1} d t + \int \frac{8 t}{t ^{2} + 1} d t

\int \frac{8 t ^{3}}{t ^{2} + 1} d t = 4 (t^{2} + 1 - ln t^{2} + 1)

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

\int \frac{8 t}{t ^{2} + 1} d t = 4 ln t^{2} + 1

= 4 (t^{2} + 1 - ln t^{2} + 1) - 3 (- arctan (t) + t) + 4 ln t^{2} + 1

簡素化

4 (t^{2} + 1 - ln t^{2} + 1) - 3 (- arctan (t) + t) + 4 ln t^{2} + 1 : 4 t^{2} - 3 t + 3 arctan (t) + 4

= 4 t^{2} - 3 t + 3 arctan (t) + 4

定数を解答に追加する

= 4 t^{2} - 3 t + 3 arctan (t) + 4 + C