What is the integral of 2x^2e^{2x} ?

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

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integral of 2x^2e^{2x}

\int 2 x^{2} e^{2 x} d x

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

Solution steps

\int 2 x^{2} e^{2 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int x^{2} e^{2 x} d x

Apply Integration By Parts

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial s} (s t)

x^{4} - 2 x^{3} + 5

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

\frac{d}{d x} (cos (3 x^{7} + 2))

\int sec (4 x) d x

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