integral of (2x^2+1)e^{x^2}

解

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

e^{x^{2}} x + C

解答ステップ

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

拡張

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

= \int 2 e^{x^{2}} x^{2} + e^{x^{2}} 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 2 e^{x^{2}} x^{2} d x + \int e^{x^{2}} d x

\int 2 e^{x^{2}} x^{2} d x = e^{x^{2}} x - \frac{π}{2} erfi (x)

\int e^{x^{2}} d x = \frac{π}{2} erfi (x)

= e^{x^{2}} x - \frac{π}{2} erfi (x) + \frac{π}{2} erfi (x)

類似した元を足す:

- \frac{π}{2} erfi (x) + \frac{π}{2} erfi (x) = 0

= e^{x^{2}} x

定数を解答に追加する

= e^{x^{2}} x + C