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

The integral of 4x^2e^{2x} is 2e^{2x}x^2-2e^{2x}x+e^{2x}+C

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integral of 4x^2e^{2x}

\int 4 x^{2} e^{2 x} d x

2 e^{2 x} x^{2} - 2 e^{2 x} x + e^{2 x} + C

Solution steps

\int 4 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

= 4 \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

= 4 (\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})

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

Simplify

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

= 2 e^{2 x} x^{2} - 2 e^{2 x} x + e^{2 x}

Add a constant to the solution

= 2 e^{2 x} x^{2} - 2 e^{2 x} x + e^{2 x} + C

\int 8 x e^{5 x} d x

x \to - \infty lim (2^{- x})

\frac{2 x}{9 x ^{2} + 1}

y = \frac{x}{x + 2}, (4, \frac{2}{6})

- 2 x^{5} + 3 x^{2} - 7

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