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

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

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

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

\frac{1}{2} e^{x^{2} - 2 x} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} e^{u}

Substitute back

u = x^{2} - 2 x

= \frac{1}{2} e^{x^{2} - 2 x}

Add a constant to the solution

= \frac{1}{2} e^{x^{2} - 2 x} + C

\frac{1}{2} t

\int (3 - 2 x)^{3} d x

\frac{\partial}{\partial x} (2 x^{2} y + sin (x^{3}))

y = 4 e^{x} + x, (0, 4)

\int_{- π}^{π} x \cdot sin (n \cdot x) d x

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