What is the integral of (x^3+2x^2+x-1)e^{2x} ?

The integral of (x^3+2x^2+x-1)e^{2x} is 1/2 e^{2x}x^3+1/4 e^{2x}x^2+1/4 e^{2x}x-5/8 e^{2x}+C

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integral of (x^3+2x^2+x-1)e^{2x}

\int (x^{3} + 2 x^{2} + x - 1) e^{2 x} d x

\frac{1}{2} e^{2 x} x^{3} + \frac{1}{4} e^{2 x} x^{2} + \frac{1}{4} e^{2 x} x - \frac{5}{8} e^{2 x} + C

Solution steps

\int (x^{3} + 2 x^{2} + x - 1) e^{2 x} d x

Expand

(x^{3} + 2 x^{2} + x - 1) e^{2 x} : e^{2 x} x^{3} + 2 e^{2 x} x^{2} + e^{2 x} x - e^{2 x}

= \int e^{2 x} x^{3} + 2 e^{2 x} x^{2} + e^{2 x} x - e^{2 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 e^{2 x} x^{3} d x + \int 2 e^{2 x} x^{2} d x + \int e^{2 x} x d x - \int e^{2 x} d x

\int e^{2 x} x^{3} d x = \frac{1}{2} e^{2 x} x^{3} - \frac{3}{2} (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}))

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

\int e^{2 x} x d x = \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

= \frac{1}{2} e^{2 x} x^{3} - \frac{3}{2} (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})) + e^{2 x} x^{2} - e^{2 x} x + \frac{1}{2} e^{2 x} + \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) - \frac{1}{2} e^{2 x}

Simplify

\frac{1}{2} e^{2 x} x^{3} - \frac{3}{2} (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})) + e^{2 x} x^{2} - e^{2 x} x + \frac{1}{2} e^{2 x} + \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) - \frac{1}{2} e^{2 x} : \frac{1}{2} e^{2 x} x^{3} + \frac{1}{4} e^{2 x} x^{2} + \frac{1}{4} e^{2 x} x - \frac{5}{8} e^{2 x}

= \frac{1}{2} e^{2 x} x^{3} + \frac{1}{4} e^{2 x} x^{2} + \frac{1}{4} e^{2 x} x - \frac{5}{8} e^{2 x}

Add a constant to the solution

= \frac{1}{2} e^{2 x} x^{3} + \frac{1}{4} e^{2 x} x^{2} + \frac{1}{4} e^{2 x} x - \frac{5}{8} e^{2 x} + C

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What is the integral of (x^3+2x^2+x-1)e^{2x} ?
The integral of (x^3+2x^2+x-1)e^{2x} is 1/2 e^{2x}x^3+1/4 e^{2x}x^2+1/4 e^{2x}x-5/8 e^{2x}+C