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 \frac{5 x - 12}{x ^{2} - 4} d x

\int \frac{e ^{12 t}}{4 e ^{6 t} - 1} d t

\int \frac{6 x ^{2} - 1}{x} d x

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

\int \frac{1}{x ^{2} 2 x - 4} d x

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