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

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

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integral of e^{t^2-t}

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

Solution steps

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

Complete the square

t^{2} - t : (t - \frac{1}{2})^{2} - \frac{1}{4}

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

Apply u-substitution

: \int e^{u^{2} - \frac{1}{4}} d u

= \int e^{u^{2} - \frac{1}{4}} d u

Simplify

Take the constant out:

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

Use the nonelementary integral:

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

Substitute back

u = t - \frac{1}{2}

Add a constant to the solution

\frac{\partial}{\partial x} (ln (5 x + 3 y))

f (z) = e^{z}

\frac{d}{d x} (1 - cos (6 x))

n = 4 \sum \infty 1

x \to \infty lim (2 x + 4)

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