What is the integral of (x^2-9)e^{2x} ?

The integral of (x^2-9)e^{2x} is 1/2 e^{2x}x^2-1/2 e^{2x}x-17/4 e^{2x}+C

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integral of (x^2-9)e^{2x}

\int (x^{2} - 9) e^{2 x} d x

\frac{1}{2} e^{2 x} x^{2} - \frac{1}{2} e^{2 x} x - \frac{17}{4} e^{2 x} + C

Solution steps

\int (x^{2} - 9) e^{2 x} d x

Apply Integration By Parts

: \frac{1}{2} e^{2 x} (x^{2} - 9) - \int e^{2 x} x d x

= \frac{1}{2} e^{2 x} (x^{2} - 9) - \int e^{2 x} x d x

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

= \frac{1}{2} e^{2 x} (x^{2} - 9) - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

Simplify

\frac{1}{2} e^{2 x} (x^{2} - 9) - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) : \frac{1}{2} e^{2 x} x^{2} - \frac{1}{2} e^{2 x} x - \frac{17}{4} e^{2 x}

= \frac{1}{2} e^{2 x} x^{2} - \frac{1}{2} e^{2 x} x - \frac{17}{4} e^{2 x}

Add a constant to the solution

= \frac{1}{2} e^{2 x} x^{2} - \frac{1}{2} e^{2 x} x - \frac{17}{4} e^{2 x} + C

\frac{d}{d x} ((2 x^{4} - 1) (5 x^{3} + 6 x))

2 sec^{2} (x)

\int \frac{x + 4}{x ^{2} + 2 x} d x

ln (x), 7

x \to - 1 + lim (4 - x)

What is the integral of (x^2-9)e^{2x} ?
The integral of (x^2-9)e^{2x} is 1/2 e^{2x}x^2-1/2 e^{2x}x-17/4 e^{2x}+C