What is the integral of e^{(x^2)/4} ?

The integral of e^{(x^2)/4} is sqrt(pi)\erfi(x/2)+C

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integral of e^{(x^2)/4}

\int e^{\frac{x ^{2}}{4}} d x

π erfi (\frac{x}{2}) + C

Solution steps

\int e^{\frac{x ^{2}}{4}} d x

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{x ^{2}}{4} = (\frac{x}{2})^{2}

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

Apply u-substitution

: \int e^{u^{2}} \cdot 2 d u

= \int e^{u^{2}} \cdot 2 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int e^{u^{2}} d u

Use the nonelementary integral:

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

= 2 \cdot \frac{π}{2} erfi (u)

Substitute back

u = \frac{x}{2}

= 2 \cdot \frac{π}{2} erfi (\frac{x}{2})

Simplify

2 \cdot \frac{π}{2} erfi (\frac{x}{2}) : π erfi (\frac{x}{2})

= π erfi (\frac{x}{2})

Add a constant to the solution

= π erfi (\frac{x}{2}) + C

x y \frac{d y}{d x} = 7

\int_{0}^{1} sin (π t + \frac{π}{4}) d t

\int 3 w d w

y^{^{'''}} + 16 y^{^{'}} = 0

y^{^{'}} = \frac{( x - 1 )}{y}

What is the integral of e^{(x^2)/4} ?
The integral of e^{(x^2)/4} is sqrt(pi)\erfi(x/2)+C