What is the integral of (1+2x^2)e^{x^2} ?

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

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

= e^{x^{2}} x

Add a constant to the solution

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

x \to 3 lim (x 13 - x^{2})

\int_{1}^{2} x d x

in v erse l a pl a ce \frac{5 e ^{- 2 s}}{s ^{2} + 4}

\frac{\partial}{\partial y} (T x^{5} y^{4} z^{8})

a re a \frac{x ^{2}}{x ^{3} - 3 x}, \frac{1}{x ^{3} - 3 x}, 2, 11

What is the integral of (1+2x^2)e^{x^2} ?
The integral of (1+2x^2)e^{x^2} is e^{x^2}x+C