integral of (1+x^2)^3

解

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

x + x^{3} + \frac{3 x ^{5}}{5} + \frac{x ^{7}}{7} + C

解答ステップ

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

拡張

(1 + x^{2})^{3} : 1 + 3 x^{2} + 3 x^{4} + x^{6}

= \int 1 + 3 x^{2} + 3 x^{4} + x^{6} 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 3 x^{2} d x + \int 3 x^{4} d x + \int x^{6} d x

\int 1 d x = x

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

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

\int x^{6} d x = \frac{x ^{7}}{7}

= x + x^{3} + \frac{3 x ^{5}}{5} + \frac{x ^{7}}{7}

定数を解答に追加する

= x + x^{3} + \frac{3 x ^{5}}{5} + \frac{x ^{7}}{7} + C