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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Simplify

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

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

y^{^{''}} - 6 y^{^{'}} + 4 y = 0

\int_{0}^{x} e^{t^{2}} d t

\int (4 + 3 x) e^{- x} d x

\frac{\partial}{\partial x} (2 x e^{x})

f (x) = 6 x + 7 x

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