integral of (1+2x^2)^2

解

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

x + \frac{4 x ^{3}}{3} + \frac{4 x ^{5}}{5} + C

解答ステップ

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

拡張

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

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

\int 1 d x = x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

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

= x + \frac{4 x ^{3}}{3} + \frac{4 x ^{5}}{5}

定数を解答に追加する

= x + \frac{4 x ^{3}}{3} + \frac{4 x ^{5}}{5} + C