What is the integral of (9-t^2)e^{t^2} ?

The integral of (9-t^2)e^{t^2} is 9*(sqrt(pi))/2 \erfi(t)-1/2 e^{t^2}t+(sqrt(pi))/4 \erfi(t)+C

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

\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

Solution steps

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

Expand

(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

Apply the Sum Rule:

\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))

Simplify

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

Add a constant to the solution

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

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What is the integral of (9-t^2)e^{t^2} ?
The integral of (9-t^2)e^{t^2} is 9*(sqrt(pi))/2 \erfi(t)-1/2 e^{t^2}t+(sqrt(pi))/4 \erfi(t)+C