integral of (4x+1)^2

解

\int (4 x + 1)^{2} d x

\frac{16 x ^{3}}{3} + 4 x^{2} + x + C

解答ステップ

\int (4 x + 1)^{2} d x

拡張

(4 x + 1)^{2} : 16 x^{2} + 8 x + 1

= \int 16 x^{2} + 8 x + 1 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^{2} d x + \int 8 x d x + \int 1 d x

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

\int 8 x d x = 4 x^{2}

\int 1 d x = x

= \frac{16 x ^{3}}{3} + 4 x^{2} + x

定数を解答に追加する

= \frac{16 x ^{3}}{3} + 4 x^{2} + x + C