What is the integral of (2x-3)e^{x^2-3x} ?

The integral of (2x-3)e^{x^2-3x} is e^{x^2-3x}+C

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integral of (2x-3)e^{x^2-3x}

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

e^{x^{2} - 3 x} + C

Solution steps

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

Apply u-substitution

= \int e^{u} d u

Use the common integral:

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

= e^{u}

Substitute back

u = x^{2} - 3 x

= e^{x^{2} - 3 x}

Add a constant to the solution

= e^{x^{2} - 3 x} + C

\frac{\partial}{\partial x} (x^{5} y)

\int \frac{4 x ^{2} + 2 x + 8}{x ( x ^{2} + 2 ) ^{2}} d x

x \to + 1 lim (- \frac{3}{x - 1})

y^{^{'}} = 2 y + x e^{x}, y (0) = - 1

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

What is the integral of (2x-3)e^{x^2-3x} ?
The integral of (2x-3)e^{x^2-3x} is e^{x^2-3x}+C