What is the integral of 8x^2e^{4x} ?

The integral of 8x^2e^{4x} is 8(1/4 e^{4x}x^2-1/32 (4e^{4x}x-e^{4x}))+C

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integral of 8x^2e^{4x}

\int 8 x^{2} e^{4 x} d x

8 (\frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x})) + C

Solution steps

\int 8 x^{2} e^{4 x} d x

Take the constant out:

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

= 8 \cdot \int x^{2} e^{4 x} d x

Apply Integration By Parts

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

= 8 (\frac{1}{4} e^{4 x} x^{2} - \int \frac{1}{2} e^{4 x} x d x)

\int \frac{1}{2} e^{4 x} x d x = \frac{1}{32} (4 e^{4 x} x - e^{4 x})

= 8 (\frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x}))

Add a constant to the solution

= 8 (\frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x})) + C

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What is the integral of 8x^2e^{4x} ?
The integral of 8x^2e^{4x} is 8(1/4 e^{4x}x^2-1/32 (4e^{4x}x-e^{4x}))+C