What is the integral of (x^2+7)e^{4x} ?

The integral of (x^2+7)e^{4x} is 1/4 e^{4x}x^2-1/8 e^{4x}x+57/32 e^{4x}+C

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integral of (x^2+7)e^{4x}

\int (x^{2} + 7) e^{4 x} d x

\frac{1}{4} e^{4 x} x^{2} - \frac{1}{8} e^{4 x} x + \frac{57}{32} e^{4 x} + C

Solution steps

\int (x^{2} + 7) e^{4 x} d x

Expand

(x^{2} + 7) e^{4 x} : e^{4 x} x^{2} + 7 e^{4 x}

= \int e^{4 x} x^{2} + 7 e^{4 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^{4 x} x^{2} d x + \int 7 e^{4 x} d x

\int e^{4 x} x^{2} d x = \frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x})

\int 7 e^{4 x} d x = \frac{7}{4} e^{4 x}

= \frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x}) + \frac{7}{4} e^{4 x}

Simplify

\frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x}) + \frac{7}{4} e^{4 x} : \frac{1}{4} e^{4 x} x^{2} - \frac{1}{8} e^{4 x} x + \frac{57}{32} e^{4 x}

= \frac{1}{4} e^{4 x} x^{2} - \frac{1}{8} e^{4 x} x + \frac{57}{32} e^{4 x}

Add a constant to the solution

= \frac{1}{4} e^{4 x} x^{2} - \frac{1}{8} e^{4 x} x + \frac{57}{32} e^{4 x} + C

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What is the integral of (x^2+7)e^{4x} ?
The integral of (x^2+7)e^{4x} is 1/4 e^{4x}x^2-1/8 e^{4x}x+57/32 e^{4x}+C