integral of (4x^2-3)^2

解

\int (4 x^{2} - 3)^{2} d x

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

解答ステップ

\int (4 x^{2} - 3)^{2} d x

拡張

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

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

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 x^{4} d x - \int 24 x^{2} d x + \int 9 d x

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

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

\int 9 d x = 9 x

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

定数を解答に追加する

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