integral of (4x^3+3)sqrt((4x^2+1))

解

\int (4 x^{3} + 3) (4 x^{2} + 1) d x

\frac{1}{30} (6 x^{2} - 1) (4 x^{2} + 1)^{\frac{3}{2}} + \frac{3}{4} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2}) + C

解答ステップ

\int (4 x^{3} + 3) 4 x^{2} + 1 d x

拡張

(4 x^{3} + 3) 4 x^{2} + 1 : 4 x^{3} 4 x^{2} + 1 + 3 4 x^{2} + 1

= \int 4 x^{3} 4 x^{2} + 1 + 3 4 x^{2} + 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 4 x^{3} 4 x^{2} + 1 d x + \int 3 4 x^{2} + 1 d x

\int 4 x^{3} 4 x^{2} + 1 d x = \frac{1}{30} (6 x^{2} - 1) (4 x^{2} + 1)^{\frac{3}{2}}

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

= \frac{1}{30} (6 x^{2} - 1) (4 x^{2} + 1)^{\frac{3}{2}} + \frac{3}{4} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2})

定数を解答に追加する

= \frac{1}{30} (6 x^{2} - 1) (4 x^{2} + 1)^{\frac{3}{2}} + \frac{3}{4} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2}) + C