integral of (9-t^2)e^{t^2}

解

\int (9 - t^{2}) e^{t^{2}} d t

9 \cdot \frac{π}{2} erfi (t) - \frac{1}{2} e^{t^{2}} t + \frac{π}{4} erfi (t) + C

解答ステップ

\int (9 - t^{2}) e^{t^{2}} d t

拡張

(9 - t^{2}) e^{t^{2}} : 9 e^{t^{2}} - e^{t^{2}} t^{2}

= \int 9 e^{t^{2}} - e^{t^{2}} t^{2} d t

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 9 e^{t^{2}} d t - \int e^{t^{2}} t^{2} d t

\int 9 e^{t^{2}} d t = 9 \cdot \frac{π}{2} erfi (t)

\int e^{t^{2}} t^{2} d t = \frac{1}{2} e^{t^{2}} t - \frac{π}{4} erfi (t)

= 9 \cdot \frac{π}{2} erfi (t) - (\frac{1}{2} e^{t^{2}} t - \frac{π}{4} erfi (t))

簡素化

= 9 \cdot \frac{π}{2} erfi (t) - \frac{1}{2} e^{t^{2}} t + \frac{π}{4} erfi (t)

定数を解答に追加する

= 9 \cdot \frac{π}{2} erfi (t) - \frac{1}{2} e^{t^{2}} t + \frac{π}{4} erfi (t) + C