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

The integral of (x-1)e^x is e^xx-2e^x+C

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integral of (x-1)e^x

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

= e^{x} (x - 1) - e^{x}

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 2 y^{^{'}} = (8 t - 7) sin (3 t)

\int e^{2 t} d t

\frac{\partial}{\partial y} (x^{3} cos (y))

x \to 0 lim (5 x^{2})

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