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

The integral of 1+e^{x^2} is x+(sqrt(pi))/2 \erfi(x)+C

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integral of 1+e^{x^2}

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

x + \frac{π}{2} erfi (x) + C

Solution steps

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

\int 1 d x = x

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

= x + \frac{π}{2} erfi (x)

Add a constant to the solution

= x + \frac{π}{2} erfi (x) + C

\frac{\partial}{\partial y} (e^{x^{2} + y^{2}} \cdot 2 y)

\int \frac{x ^{\frac{1}{3}} - 3}{x ^{\frac{2}{3}}} d x

\int_{0}^{2} 1 + 4 x^{2} d x

q^{^{'}} + \frac{q}{20 + t} = 1, q (0) = 5

x^{8} (x - 3)^{7}

What is the integral of 1+e^{x^2} ?
The integral of 1+e^{x^2} is x+(sqrt(pi))/2 \erfi(x)+C