What is the integral of (3x-4)e-x ?

The integral of (3x-4)e-x is e((3x^2)/2-4x)-(x^2)/2+C

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integral of (3x-4)e-x

\int (3 x - 4) e - x d x

e (\frac{3 x ^{2}}{2} - 4 x) - \frac{x ^{2}}{2} + C

Solution steps

\int (3 x - 4) e - x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int (3 x - 4) e d x - \int x d x

\int (3 x - 4) e d x = e (\frac{3 x ^{2}}{2} - 4 x)

\int x d x = \frac{x ^{2}}{2}

= e (\frac{3 x ^{2}}{2} - 4 x) - \frac{x ^{2}}{2}

Add a constant to the solution

= e (\frac{3 x ^{2}}{2} - 4 x) - \frac{x ^{2}}{2} + C

\frac{\partial}{\partial x} (\frac{y ^{2}}{2})

\frac{d}{d x} (\frac{- 9 x - 1}{- 6 x + 2})

\frac{\partial}{\partial y} (40 x^{\frac{4}{5}} y^{\frac{1}{5}})

\frac{d y}{d t} + y \cdot cot (t) = cos (t)

\frac{d}{d x} ((3 x^{2} - 6 x + 6) e^{x})

What is the integral of (3x-4)e-x ?
The integral of (3x-4)e-x is e((3x^2)/2-4x)-(x^2)/2+C