integral of (4t^2+3)^2

解

\int (4 t^{2} + 3)^{2} d t

\frac{16 t ^{5}}{5} + 8 t^{3} + 9 t + C

解答ステップ

\int (4 t^{2} + 3)^{2} d t

拡張

(4 t^{2} + 3)^{2} : 16 t^{4} + 24 t^{2} + 9

= \int 16 t^{4} + 24 t^{2} + 9 d t

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 16 t^{4} d t + \int 24 t^{2} d t + \int 9 d t

\int 16 t^{4} d t = \frac{16 t ^{5}}{5}

\int 24 t^{2} d t = 8 t^{3}

\int 9 d t = 9 t

= \frac{16 t ^{5}}{5} + 8 t^{3} + 9 t

定数を解答に追加する

= \frac{16 t ^{5}}{5} + 8 t^{3} + 9 t + C