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

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

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integral of e^{x^2-2x}

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

\frac{π}{2 e} erfi (x - 1) + C

Solution steps

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

Complete the square

x^{2} - 2 x : (x - 1)^{2} - 1

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

Apply u-substitution

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

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

Simplify

e^{u^{2} - 1} : e^{u^{2}} e^{- 1}

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

Take the constant out:

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

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Use the nonelementary integral:

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

= \frac{1}{e} \cdot \frac{π}{2} erfi (u)

Substitute back

u = x - 1

= \frac{1}{e} \cdot \frac{π}{2} erfi (x - 1)

Simplify

\frac{1}{e} \cdot \frac{π}{2} erfi (x - 1) : \frac{π}{2 e} erfi (x - 1)

= \frac{π}{2 e} erfi (x - 1)

Add a constant to the solution

= \frac{π}{2 e} erfi (x - 1) + C

\frac{d}{d x} (x^{\frac{1}{2}} \cdot csc (x) sin (x))

2 y^{^{'}} - 6 x^{2} = 0

\frac{\partial}{\partial x} (y sin (x))

\int_{0}^{3} (5 t - 8) d t

e^{2 x^{2}}

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