integral of (4t^3-t^2+16t)/(t^2+4)

解

\int \frac{4 t ^{3} - t ^{2} + 16 t}{t ^{2} + 4} d t

2 t^{2} - t + 2 arctan (\frac{t}{2}) + 8 + C

解答ステップ

\int \frac{4 t ^{3} - t ^{2} + 16 t}{t ^{2} + 4} d t

拡張

\frac{4 t ^{3} - t ^{2} + 16 t}{t ^{2} + 4} : \frac{4 t ^{3}}{t ^{2} + 4} - \frac{t ^{2}}{t ^{2} + 4} + \frac{16 t}{t ^{2} + 4}

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} + 4} d t - \int \frac{t ^{2}}{t ^{2} + 4} d t + \int \frac{16 t}{t ^{2} + 4} d t

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

\int \frac{t ^{2}}{t ^{2} + 4} d t = - 2 arctan (\frac{t}{2}) + t

\int \frac{16 t}{t ^{2} + 4} d t = 8 ln t^{2} + 4

= 2 (t^{2} + 4 - 4 ln t^{2} + 4) - (- 2 arctan (\frac{t}{2}) + t) + 8 ln t^{2} + 4

簡素化

2 (t^{2} + 4 - 4 ln t^{2} + 4) - (- 2 arctan (\frac{t}{2}) + t) + 8 ln t^{2} + 4 : 2 t^{2} - t + 2 arctan (\frac{t}{2}) + 8

= 2 t^{2} - t + 2 arctan (\frac{t}{2}) + 8

定数を解答に追加する

= 2 t^{2} - t + 2 arctan (\frac{t}{2}) + 8 + C