What is the integral of (x^3-3x^2+4)e^{2-3x} ?

The integral of (x^3-3x^2+4)e^{2-3x} is

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integral of (x^3-3x^2+4)e^{2-3x}

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

\frac{1}{81} (- 90 e^{- 3 x + 2} x + e^{- 3 x + 2} (- 3 x + 2)^{3} - 9 e^{- 3 x + 2} (- 3 x + 2)^{2} + 22 e^{- 3 x + 2}) + \frac{1}{9} (9 e^{- 3 x + 2} x^{2} + 6 e^{- 3 x + 2} x + 2 e^{- 3 x + 2}) - \frac{4}{3} e^{2 - 3 x} + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} ( u - 2 ) ^{3}}{81} + \frac{e ^{u} ( u - 2 ) ^{2}}{9} - \frac{4}{3} e^{u} d u

= \int \frac{e ^{u} ( u - 2 ) ^{3}}{81} + \frac{e ^{u} ( u - 2 ) ^{2}}{9} - \frac{4}{3} e^{u} d u

Apply the Sum Rule:

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

= \int \frac{e ^{u} ( u - 2 ) ^{3}}{81} d u + \int \frac{e ^{u} ( u - 2 ) ^{2}}{9} d u - \int \frac{4}{3} e^{u} d u

\int \frac{e ^{u} ( u - 2 ) ^{3}}{81} d u = \frac{1}{81} (e^{u} u^{3} - 9 e^{u} u^{2} + 30 e^{u} u - 38 e^{u})

\int \frac{e ^{u} ( u - 2 ) ^{2}}{9} d u = \frac{1}{9} (e^{u} u^{2} - 6 e^{u} u + 10 e^{u})

\int \frac{4}{3} e^{u} d u = \frac{4}{3} e^{u}

= \frac{1}{81} (e^{u} u^{3} - 9 e^{u} u^{2} + 30 e^{u} u - 38 e^{u}) + \frac{1}{9} (e^{u} u^{2} - 6 e^{u} u + 10 e^{u}) - \frac{4}{3} e^{u}

Substitute back

u = 2 - 3 x

= \frac{1}{81} (e^{2 - 3 x} (2 - 3 x)^{3} - 9 e^{2 - 3 x} (2 - 3 x)^{2} + 30 e^{2 - 3 x} (2 - 3 x) - 38 e^{2 - 3 x}) + \frac{1}{9} (e^{2 - 3 x} (2 - 3 x)^{2} - 6 e^{2 - 3 x} (2 - 3 x) + 10 e^{2 - 3 x}) - \frac{4}{3} e^{2 - 3 x}

Simplify

\frac{1}{81} (e^{2 - 3 x} (2 - 3 x)^{3} - 9 e^{2 - 3 x} (2 - 3 x)^{2} + 30 e^{2 - 3 x} (2 - 3 x) - 38 e^{2 - 3 x}) + \frac{1}{9} (e^{2 - 3 x} (2 - 3 x)^{2} - 6 e^{2 - 3 x} (2 - 3 x) + 10 e^{2 - 3 x}) - \frac{4}{3} e^{2 - 3 x} : \frac{1}{81} (- 90 e^{- 3 x + 2} x + e^{- 3 x + 2} (- 3 x + 2)^{3} - 9 e^{- 3 x + 2} (- 3 x + 2)^{2} + 22 e^{- 3 x + 2}) + \frac{1}{9} (9 e^{- 3 x + 2} x^{2} + 6 e^{- 3 x + 2} x + 2 e^{- 3 x + 2}) - \frac{4}{3} e^{2 - 3 x}

= \frac{1}{81} (- 90 e^{- 3 x + 2} x + e^{- 3 x + 2} (- 3 x + 2)^{3} - 9 e^{- 3 x + 2} (- 3 x + 2)^{2} + 22 e^{- 3 x + 2}) + \frac{1}{9} (9 e^{- 3 x + 2} x^{2} + 6 e^{- 3 x + 2} x + 2 e^{- 3 x + 2}) - \frac{4}{3} e^{2 - 3 x}

Add a constant to the solution

= \frac{1}{81} (- 90 e^{- 3 x + 2} x + e^{- 3 x + 2} (- 3 x + 2)^{3} - 9 e^{- 3 x + 2} (- 3 x + 2)^{2} + 22 e^{- 3 x + 2}) + \frac{1}{9} (9 e^{- 3 x + 2} x^{2} + 6 e^{- 3 x + 2} x + 2 e^{- 3 x + 2}) - \frac{4}{3} e^{2 - 3 x} + C

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