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

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

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

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

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

Solution steps

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

Expand

(x^{4} + x) e^{2 x} : e^{2 x} x^{4} + e^{2 x} x

= \int e^{2 x} x^{4} + e^{2 x} 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^{4} d x + \int e^{2 x} x d x

\int e^{2 x} x^{4} d x = \frac{1}{4} e^{2 x} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3)

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

= \frac{1}{4} e^{2 x} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

Simplify

\frac{1}{4} e^{2 x} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) : \frac{1}{2} e^{2 x} x^{4} - e^{2 x} x^{3} + \frac{3}{2} e^{2 x} x^{2} - e^{2 x} x + \frac{1}{2} e^{2 x}

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

Add a constant to the solution

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

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