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

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

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

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

\frac{1}{2} e^{2 x} x^{2} - \frac{1}{2} e^{2 x} x - \frac{7}{4} e^{2 x} + C

Solution steps

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

Apply Integration By Parts

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

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

Simplify

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

= \frac{1}{2} e^{2 x} x^{2} - \frac{1}{2} e^{2 x} x - \frac{7}{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{7}{4} e^{2 x} + C

\frac{d}{d x} (\frac{2 x}{6 x ^{2} + 1})

\frac{d y}{d x} = x y + x

\frac{d}{d x} (x - 3 ln (x))

\frac{d}{d x} (e^{- x t})

x \to 3 + lim (\frac{[ x ]}{x})

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