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

The integral of (x-1)^2(e^{x+1}-1) is -(x^3)/3+e^{x+1}x^2+x^2-4e^{x+1}x-x+5e^{x+1}+C

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

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

- \frac{x ^{3}}{3} + e^{x + 1} x^{2} + x^{2} - 4 e^{x + 1} x - x + 5 e^{x + 1} + C

Solution steps

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

Expand

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

= \int e^{x + 1} x^{2} - x^{2} - 2 e^{x + 1} x + 2 x + e^{x + 1} - 1 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 e^{x + 1} x^{2} d x - \int x^{2} d x - \int 2 e^{x + 1} x d x + \int 2 x d x + \int e^{x + 1} d x - \int 1 d x

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

\int x^{2} d x = \frac{x ^{3}}{3}

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

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

\int e^{x + 1} d x = e^{1 + x}

\int 1 d x = x

= x^{2} e^{1 + x} - 2 (e^{1 + x} x - e^{1 + x}) - \frac{x ^{3}}{3} - 2 (e^{1 + x} x - e^{1 + x}) + x^{2} + e^{1 + x} - x

Simplify

x^{2} e^{1 + x} - 2 (e^{1 + x} x - e^{1 + x}) - \frac{x ^{3}}{3} - 2 (e^{1 + x} x - e^{1 + x}) + x^{2} + e^{1 + x} - x : - \frac{x ^{3}}{3} + e^{x + 1} x^{2} + x^{2} - 4 e^{x + 1} x - x + 5 e^{x + 1}

= - \frac{x ^{3}}{3} + e^{x + 1} x^{2} + x^{2} - 4 e^{x + 1} x - x + 5 e^{x + 1}

Add a constant to the solution

= - \frac{x ^{3}}{3} + e^{x + 1} x^{2} + x^{2} - 4 e^{x + 1} x - x + 5 e^{x + 1} + C

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What is the integral of (x-1)^2(e^{x+1}-1) ?
The integral of (x-1)^2(e^{x+1}-1) is -(x^3)/3+e^{x+1}x^2+x^2-4e^{x+1}x-x+5e^{x+1}+C