integral of xTg(x^2+a)

解

\int x T g (x^{2} + a) d x

g T (\frac{a x ^{2}}{2} + \frac{x ^{4}}{4}) + C

解答ステップ

\int x T g (x^{2} + a) d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= g T \cdot \int (a + x^{2}) x d x

拡張

(a + x^{2}) x : a x + x^{3}

= g T \cdot \int a x + x^{3} 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

= g T (\int a x d x + \int x^{3} d x)

\int a x d x = \frac{a x ^{2}}{2}

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

= g T (\frac{a x ^{2}}{2} + \frac{x ^{4}}{4})

定数を解答に追加する

= g T (\frac{a x ^{2}}{2} + \frac{x ^{4}}{4}) + C